高等学校数学III 積分法/演習問題

次の不定積分を計算せよ。
以下の解答においては、問iにおいて計算すべき積分をIiとおくことにする。
I 1 = ∫ ( x + 2 x ) 2 d x = ∫ ( 1 + 4 x + 4 x 2 ) d x = x + 4 log  | x | − 4 x + C {\displaystyle {\begin{aligned}I_{1}&=\int \left({\frac {x+2}{x}}\right)^{2}dx\\&=\int \left(1+{\frac {4}{x}}+{\frac {4}{x^{2}}}\right)dx\\&=x+4\log |x|-{\frac {4}{x}}+C\end{aligned}}}
I 2 = ∫ ( x + 1 ) ( x − 2 ) d x = ∫ ( x − x − 2 ) d x = x 2 2 − 2 3 x x − 2 x + C {\displaystyle {\begin{aligned}I_{2}&=\int \left({\sqrt {x}}+1\right)\left({\sqrt {x}}-2\right)dx\\&=\int \left(x-{\sqrt {x}}-2\right)dx\\&={\frac {x^{2}}{2}}-{\frac {2}{3}}x{\sqrt {x}}-2x+C\\\end{aligned}}}
I 3 = ∫ ( 2 x + 3 ) 2 d x = ∫ ( 2 x + 6 2 x + 9 ) d x = x 2 + 4 x 2 x + 9 x + C {\displaystyle {\begin{aligned}I_{3}&=\int \left({\sqrt {2x}}+3\right)^{2}dx\\&=\int \left(2x+6{\sqrt {2x}}+9\right)dx\\&=x^{2}+4x{\sqrt {2x}}+9x+C\end{aligned}}}
I 4 = ∫ x − cos 2  x x cos 2  x d x = ∫ ( 1 cos 2  x − 1 x ) d x = tan  x − log  | x | + C {\displaystyle {\begin{aligned}I_{4}&=\int {\frac {x-\cos ^{2}x}{x\cos ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}-{\frac {1}{x}}\right)dx\\&=\tan x-\log |x|+C\end{aligned}}}
I 5 = ∫ ( tan 2  x + 1 tan 2  x ) d x = ∫ ( ( 1 cos 2  x − 1 ) + ( 1 sin 2  x − 1 ) ) d x = tan  x − 1 tan  x − 2 x + C {\displaystyle {\begin{aligned}I_{5}&=\int \left(\tan ^{2}x+{\frac {1}{\tan ^{2}x}}\right)dx\\&=\int \left(\left({\frac {1}{\cos ^{2}x}}-1\right)+\left({\frac {1}{\sin ^{2}x}}-1\right)\right)dx\\&=\tan x-{\frac {1}{\tan x}}-2x+C\end{aligned}}}
I 6 = ∫ ( e x + 1 x ) d x = e x + log  | x | + C {\displaystyle {\begin{aligned}I_{6}&=\int \left(e^{x}+{\frac {1}{x}}\right)dx\\&=e^{x}+\log |x|+C\end{aligned}}}
I 7 = ∫ 3 x + 2 d x = 3 x + 2 log  3 + C {\displaystyle {\begin{aligned}I_{7}&=\int 3^{x+2}dx\\&={\frac {3^{x+2}}{\log 3}}+C\end{aligned}}}
I 8 = ∫ ( x + 2 ) ( x 2 + 4 x + 1 ) 3 d x {\displaystyle {\begin{aligned}I_{8}&=\int (x+2)(x^{2}+4x+1)^{3}dx\\\end{aligned}}}
t = x 2 + 4 x + 1 {\displaystyle t=x^{2}+4x+1} とおくと d t = ( 2 x + 4 ) d x = 2 ( x + 2 ) d x {\displaystyle dt=(2x+4)dx=2(x+2)dx} なので、
I 8 = ∫ t 3 2 d t = t 4 8 + C = 1 8 ( x 2 + 4 x + 1 ) 4 + C {\displaystyle {\begin{aligned}I_{8}&=\int {\frac {t^{3}}{2}}dt\\&={\frac {t^{4}}{8}}+C\\&={\frac {1}{8}}(x^{2}+4x+1)^{4}+C\end{aligned}}}
I 9 = ∫ x + 1 ( 3 x − 1 ) 3 d x = ∫ ( 1 3 ( 3 x − 1 ) 2 + 4 3 ( 3 x − 1 ) 3 ) d x = − 1 9 ( 3 x − 1 ) − 2 9 ( 3 x − 1 ) 2 + C = − 3 x + 1 9 ( 3 x − 1 ) 2 + C {\displaystyle {\begin{aligned}I_{9}&=\int {\frac {x+1}{(3x-1)^{3}}}dx\\&=\int \left({\frac {1}{3(3x-1)^{2}}}+{\frac {4}{3(3x-1)^{3}}}\right)dx\\&=-{\frac {1}{9(3x-1)}}-{\frac {2}{9(3x-1)^{2}}}+C\\&=-{\frac {3x+1}{9(3x-1)^{2}}}+C\end{aligned}}}
別解
t = 3 x − 1 {\displaystyle t=3x-1} と置く。 x = t + 1 3 , x + 1 = t + 4 3 , d t d x = 3 , d t = 3 d x {\displaystyle x={\frac {t+1}{3}},\;\;x+1={\frac {t+4}{3}},\;\;{\frac {dt}{dx}}=3,\;\;dt=3dx} なので、
I 9 = 1 9 ∫ ( t + 4 t 3 ) d t = 1 9 ∫ ( 1 t 2 + 4 t 3 ) d t = − 1 9 ( 1 t + 2 t 2 ) + C = − 1 9 ( 1 3 x − 1 + 2 ( 3 x − 1 ) 2 ) + C = − 3 x + 1 9 ( 3 x − 1 ) 2 + C {\displaystyle {\begin{aligned}I_{9}&={\frac {1}{9}}\int \left({\frac {t+4}{t^{3}}}\right)dt\\&={\frac {1}{9}}\int \left({\frac {1}{t^{2}}}+{\frac {4}{t^{3}}}\right)dt\\&=-{\frac {1}{9}}\left({\frac {1}{t}}+{\frac {2}{t^{2}}}\right)+C\\&=-{\frac {1}{9}}\left({\frac {1}{3x-1}}+{\frac {2}{(3x-1)^{2}}}\right)+C\\&=-{\frac {3x+1}{9(3x-1)^{2}}}+C\end{aligned}}}
I 10 = ∫ cos 3  x ⋅ sin 2  x d x = ∫ cos  x ( 1 − sin 2  x ) sin 2  x d x = ∫ cos  x ( sin 2  x − sin 4  x ) d x {\displaystyle {\begin{aligned}I_{10}&=\int \cos ^{3}x\cdot \sin ^{2}xdx\\&=\int \cos x(1-\sin ^{2}x)\sin ^{2}xdx\\&=\int \cos x(\sin ^{2}x-\sin ^{4}x)dx\\\end{aligned}}}
t = sin  x {\displaystyle t=\sin x} とおくと d t = cos  x d x {\displaystyle dt=\cos xdx} なので、
I 10 = ∫ ( t 2 − t 4 ) d t = t 3 3 − t 5 5 + C = 1 3 sin 3  x − 1 5 sin 5  x + C {\displaystyle {\begin{aligned}I_{10}&=\int (t^{2}-t^{4})dt\\&={\frac {t^{3}}{3}}-{\frac {t^{5}}{5}}+C\\&={\frac {1}{3}}\sin ^{3}x-{\frac {1}{5}}\sin ^{5}x+C\end{aligned}}}
I 11 = ∫ d x ( 1 + tan  x ) cos 2  x {\displaystyle {\begin{aligned}I_{11}&=\int {\frac {dx}{(1+\tan x)\cos ^{2}x}}\\\end{aligned}}}
t = tan  x {\displaystyle t=\tan x} とおくと d t = d x cos 2  x {\displaystyle dt={\frac {dx}{\cos ^{2}x}}} なので、
I 11 = ∫ d t 1 + t = log  | 1 + t | + C = log  | 1 + tan  x | + C {\displaystyle {\begin{aligned}I_{11}&=\int {\frac {dt}{1+t}}\\&=\log |1+t|+C\\&=\log |1+\tan x|+C\end{aligned}}}
I 12 = ∫ ( 2 x + 1 ) e x 2 + x + 5 d x {\displaystyle {\begin{aligned}I_{12}&=\int (2x+1)e^{x^{2}+x+5}dx\\\end{aligned}}}
t = x 2 + x + 5 {\displaystyle t=x^{2}+x+5} とおくと d t = ( 2 x + 1 ) d x {\displaystyle dt=(2x+1)dx} なので、
I 12 = ∫ e t d t = e t + C = e x 2 + x + 5 + C {\displaystyle {\begin{aligned}I_{12}&=\int e^{t}dt\\&=e^{t}+C\\&=e^{x^{2}+x+5}+C\end{aligned}}}
I 13 = ∫ e 4 x e 2 x + 1 d x {\displaystyle {\begin{aligned}I_{13}&=\int {\frac {e^{4x}}{e^{2x}+1}}dx\\\end{aligned}}}
t = e 2 x + 1 {\displaystyle t=e^{2x}+1} とおくと d t = 2 e 2 x d x {\displaystyle dt=2e^{2x}dx} なので、
I 13 = ∫ t − 1 2 t d t = 1 2 ∫ ( 1 − 1 t ) d t = 1 2 ( t − log  t ) + C ′ = 1 2 ( e 2 x + 1 − log  ( e 2 x + 1 ) ) + C ′ = 1 2 ( e 2 x − log  ( e 2 x + 1 ) ) + C {\displaystyle {\begin{aligned}I_{13}&=\int {\frac {t-1}{2t}}dt\\&={\frac {1}{2}}\int \left(1-{\frac {1}{t}}\right)dt\\&={\frac {1}{2}}(t-\log t)+C'\\&={\frac {1}{2}}\left(e^{2x}+1-\log(e^{2x}+1)\right)+C'\\&={\frac {1}{2}}\left(e^{2x}-\log(e^{2x}+1)\right)+C\end{aligned}}}
I 14 = ∫ x 1 − x d x {\displaystyle {\begin{aligned}I_{14}&=\int x{\sqrt {1-x}}dx\\\end{aligned}}}
t = 1 − x {\displaystyle t={\sqrt {1-x}}} とおくと、 t 2 = 1 − x , x = 1 − t 2 , d x = − 2 t d t {\displaystyle t^{2}=1-x,\;\;x=1-t^{2},\;\;dx=-2tdt} なので、
I 14 = − 2 ∫ ( 1 − t 2 ) ⋅ t ⋅ t d t = − 2 ∫ ( t 2 − t 4 ) d t = − 2 ( t 3 3 − t 5 5 ) + C = − 2 t 3 15 ( 5 − 3 t 2 ) + C = − 2 15 ( 1 − x ) 1 − x ( 5 − 3 ( 1 − x ) ) + C = − 2 15 ( 3 x + 2 ) ( 1 − x ) 1 − x + C {\displaystyle {\begin{aligned}I_{14}&=-2\int (1-t^{2})\cdot t\cdot tdt\\&=-2\int (t^{2}-t^{4})dt\\&=-2\left({\frac {t^{3}}{3}}-{\frac {t^{5}}{5}}\right)+C\\&=-{\frac {2t^{3}}{15}}(5-3t^{2})+C\\&=-{\frac {2}{15}}(1-x){\sqrt {1-x}}(5-3(1-x))+C\\&=-{\frac {2}{15}}(3x+2)(1-x){\sqrt {1-x}}+C\\\end{aligned}}}
I 15 = ∫ 3 x − 1 x + 1 d x {\displaystyle {\begin{aligned}I_{15}&=\int {\frac {3x-1}{\sqrt {x+1}}}dx\\\end{aligned}}}
t = x + 1 {\displaystyle t={\sqrt {x+1}}} とおくと、 t 2 = x + 1 , x = t 2 − 1 , d x = 2 t d t {\displaystyle t^{2}=x+1,\;\;x=t^{2}-1,\;\;dx=2tdt} なので、
I 15 = ∫ 3 ( t 2 − 1 ) − 1 t ⋅ 2 t d t = 2 ∫ ( 3 t 2 − 4 ) d t = 2 ( t 3 − 4 t ) + C = 2 t ( t 2 − 4 ) + C = 2 ( ( x + 1 ) − 4 ) x + 1 + C = 2 ( x − 3 ) x + 1 + C {\displaystyle {\begin{aligned}I_{15}&=\int {\frac {3(t^{2}-1)-1}{t}}\cdot 2tdt\\&=2\int (3t^{2}-4)dt\\&=2(t^{3}-4t)+C\\&=2t(t^{2}-4)+C\\&=2((x+1)-4){\sqrt {x+1}}+C\\&=2(x-3){\sqrt {x+1}}+C\\\end{aligned}}}
I 16 = ∫ x ( x 2 + 4 ) 2 d x {\displaystyle {\begin{aligned}I_{16}&=\int {\frac {x}{(x^{2}+4)^{2}}}dx\\\end{aligned}}}
t = x 2 + 4 {\displaystyle t=x^{2}+4} とおくと d t = 2 x d x {\displaystyle dt=2xdx} なので、
I 16 = ∫ d t 2 t 2 = − 1 2 t + C = − 1 2 ( x 2 + 4 ) + C {\displaystyle {\begin{aligned}I_{16}&=\int {\frac {dt}{2t^{2}}}\\&=-{\frac {1}{2t}}+C\\&=-{\frac {1}{2(x^{2}+4)}}+C\end{aligned}}}
I 17 = ∫ x 2 + 1 x 4 − 5 x 2 + 4 d x = 1 3 ∫ ( 5 x 2 − 4 − 2 x 2 − 1 ) d x = 1 3 ∫ ( 5 4 ( 1 x − 2 − 1 x + 2 ) − ( 1 x − 1 − 1 x + 1 ) ) d x = 5 12 log  | x − 2 x + 2 | − 1 3 log  | x − 1 x + 1 | + C = 1 3 log  | x + 1 x − 1 | − 5 12 log  | x + 2 x − 2 | + C {\displaystyle {\begin{aligned}I_{17}&=\int {\frac {x^{2}+1}{x^{4}-5x^{2}+4}}dx\\&={\frac {1}{3}}\int \left({\frac {5}{x^{2}-4}}-{\frac {2}{x^{2}-1}}\right)dx\\&={\frac {1}{3}}\int \left({\frac {5}{4}}\left({\frac {1}{x-2}}-{\frac {1}{x+2}}\right)-\left({\frac {1}{x-1}}-{\frac {1}{x+1}}\right)\right)dx\\&={\frac {5}{12}}\log \left|{\frac {x-2}{x+2}}\right|-{\frac {1}{3}}\log \left|{\frac {x-1}{x+1}}\right|+C\\&={\frac {1}{3}}\log \left|{\frac {x+1}{x-1}}\right|-{\frac {5}{12}}\log \left|{\frac {x+2}{x-2}}\right|+C\\\end{aligned}}}
I 18 = ∫ 3 x + 2 x ( x + 1 ) 3 d x = ∫ ( 1 ( x + 1 ) 3 + 2 x ( x + 1 ) 2 ) d x = ∫ ( 1 ( x + 1 ) 3 − 2 ( x + 1 ) 2 + 2 x ( x + 1 ) ) d x = ∫ ( 1 ( x + 1 ) 3 − 2 ( x + 1 ) 2 + 2 x − 2 x + 1 ) d x = − 1 2 ( x + 1 ) 2 + 2 x + 1 + 2 log  | x x + 1 | + C = 2 log  | x x + 1 | + 2 x + 1 − 1 2 ( x + 1 ) 2 + C {\displaystyle {\begin{aligned}I_{18}&=\int {\frac {3x+2}{x(x+1)^{3}}}dx\\&=\int \left({\frac {1}{(x+1)^{3}}}+{\frac {2}{x(x+1)^{2}}}\right)dx\\&=\int \left({\frac {1}{(x+1)^{3}}}-{\frac {2}{(x+1)^{2}}}+{\frac {2}{x(x+1)}}\right)dx\\&=\int \left({\frac {1}{(x+1)^{3}}}-{\frac {2}{(x+1)^{2}}}+{\frac {2}{x}}-{\frac {2}{x+1}}\right)dx\\&=-{\frac {1}{2(x+1)^{2}}}+{\frac {2}{x+1}}+2\log \left|{\frac {x}{x+1}}\right|+C\\&=2\log \left|{\frac {x}{x+1}}\right|+{\frac {2}{x+1}}-{\frac {1}{2(x+1)^{2}}}+C\end{aligned}}}
I 19 = ∫ x x + 1 3 − 1 d x = ∫ x ( ( x + 1 3 ) 2 + x + 1 3 + 1 ) x d x = ∫ ( ( x + 1 ) 2 3 + ( x + 1 ) 1 3 + 1 ) d x = 3 5 ( x + 1 ) 5 3 + 3 4 ( x + 1 ) 4 3 + x + C {\displaystyle {\begin{aligned}I_{19}&=\int {\frac {x}{{\sqrt[{3}]{x+1}}-1}}dx\\&=\int {\frac {x\left(\left({\sqrt[{3}]{x+1}}\right)^{2}+{\sqrt[{3}]{x+1}}+1\right)}{x}}dx\\&=\int \left((x+1)^{\frac {2}{3}}+(x+1)^{\frac {1}{3}}+1\right)dx\\&={\frac {3}{5}}(x+1)^{\frac {5}{3}}+{\frac {3}{4}}(x+1)^{\frac {4}{3}}+x+C\\\end{aligned}}}
I 20 = ∫ cos  x sin  x ( sin  x + 1 ) d x {\displaystyle {\begin{aligned}I_{20}&=\int {\frac {\cos x}{\sin x(\sin x+1)}}dx\\\end{aligned}}}
t = sin  x {\displaystyle t=\sin x} とおくと d t = cos  x d x {\displaystyle dt=\cos xdx} なので、
I 20 = ∫ d t t ( t + 1 ) = ∫ ( 1 t − 1 t + 1 ) d t = log  | t t + 1 | + C = log  | sin  x sin  x + 1 | + C {\displaystyle {\begin{aligned}I_{20}&=\int {\frac {dt}{t(t+1)}}\\&=\int \left({\frac {1}{t}}-{\frac {1}{t+1}}\right)dt\\&=\log \left|{\frac {t}{t+1}}\right|+C\\&=\log \left|{\frac {\sin x}{\sin x+1}}\right|+C\\\end{aligned}}}
I 21 = ∫ ( tan  x + 1 tan  x ) 2 d x = ∫ ( tan 2  x + 2 + 1 tan 2  x ) d x = ∫ ( ( 1 cos 2  x − 1 ) + 2 + ( 1 sin 2  x − 1 ) ) d x = tan  x − 1 tan  x + C {\displaystyle {\begin{aligned}I_{21}&=\int \left(\tan x+{\frac {1}{\tan x}}\right)^{2}dx\\&=\int \left(\tan ^{2}x+2+{\frac {1}{\tan ^{2}x}}\right)dx\\&=\int \left(\left({\frac {1}{\cos ^{2}x}}-1\right)+2+\left({\frac {1}{\sin ^{2}x}}-1\right)\right)dx\\&=\tan x-{\frac {1}{\tan x}}+C\\\end{aligned}}}
I 22 = ∫ tan 4  x d x = ∫ tan 2  x tan 2  x d x = ∫ tan 2  x ( 1 cos 2  x − 1 ) d x = ∫ ( tan 2  x cos 2  x − tan 2  x ) d x = ∫ ( tan 2  x cos 2  x − ( 1 cos 2  x − 1 ) ) d x = ∫ tan 2  x cos 2  x d x − ( tan  x − x ) {\displaystyle {\begin{aligned}I_{22}&=\int \tan ^{4}xdx\\&=\int \tan ^{2}x\tan ^{2}xdx\\&=\int \tan ^{2}x\left({\frac {1}{\cos ^{2}x}}-1\right)dx\\&=\int \left({\frac {\tan ^{2}x}{\cos ^{2}x}}-\tan ^{2}x\right)dx\\&=\int \left({\frac {\tan ^{2}x}{\cos ^{2}x}}-\left({\frac {1}{\cos ^{2}x}}-1\right)\right)dx\\&=\int {\frac {\tan ^{2}x}{\cos ^{2}x}}dx-(\tan x-x)\\\end{aligned}}}
t = tan  x {\displaystyle t=\tan x} とおくと d t = 1 cos 2  x d x {\displaystyle dt={\frac {1}{\cos ^{2}x}}dx} なので、
I 22 = ∫ t 2 d t − ( tan  x − x ) = 1 3 t 3 − tan  x + x + C = 1 3 tan 3  x − tan  x + x + C {\displaystyle {\begin{aligned}I_{22}&=\int t^{2}dt-(\tan x-x)\\&={\frac {1}{3}}t^{3}-\tan x+x+C\\&={\frac {1}{3}}\tan ^{3}x-\tan x+x+C\end{aligned}}}
I 23 = ∫ sin  2 x 1 + sin  x d x = ∫ 2 sin  x cos  x 1 + sin  x d x {\displaystyle {\begin{aligned}I_{23}&=\int {\frac {\sin 2x}{1+\sin x}}dx\\&=\int {\frac {2\sin x\cos x}{1+\sin x}}dx\end{aligned}}}
t = sin  x {\displaystyle t=\sin x} とおくと d t = cos  x d x {\displaystyle dt=\cos xdx} なので、
I 23 = ∫ 2 t 1 + t d t = 2 ∫ ( 1 − 1 1 + t ) d t = 2 t − 2 log  | 1 + t | + C = 2 sin  x − 2 log  ( 1 + sin  x ) + C {\displaystyle {\begin{aligned}I_{23}&=\int {\frac {2t}{1+t}}dt\\&=2\int \left(1-{\frac {1}{1+t}}\right)dt\\&=2t-2\log |1+t|+C\\&=2\sin x-2\log \left(1+\sin x\right)+C\end{aligned}}}
I 24 = ∫ x 1 − cos  x d x = ∫ x 2 sin 2  x 2 d x = − x tan  x 2 + ∫ 1 tan  x 2 d x = − x tan  x 2 + ∫ cos  x 2 sin  x 2 d x {\displaystyle {\begin{aligned}I_{24}&=\int {\frac {x}{1-\cos x}}dx\\&=\int {\frac {x}{2\sin ^{2}{\frac {x}{2}}}}dx\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+\int {\frac {1}{\tan {\frac {x}{2}}}}dx\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+\int {\frac {\cos {\frac {x}{2}}}{\sin {\frac {x}{2}}}}dx\end{aligned}}}
t = sin  x 2 {\displaystyle t=\sin {\frac {x}{2}}} とおくと d t = 1 2 cos  x 2 d x {\displaystyle dt={\frac {1}{2}}\cos {\frac {x}{2}}dx} なので、
I 24 = − x tan  x 2 + 2 ∫ d t t = − x tan  x 2 + 2 log  | t | + C = − x tan  x 2 + 2 log  | sin  x 2 | + C {\displaystyle {\begin{aligned}I_{24}&=-{\frac {x}{\tan {\frac {x}{2}}}}+2\int {\frac {dt}{t}}\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+2\log |t|+C\\&=-{\frac {x}{\tan {\frac {x}{2}}}}+2\log \left|\sin {\frac {x}{2}}\right|+C\end{aligned}}}
I 25 = ∫ 1 1 − sin  x d x = ∫ 1 + sin  x 1 − sin 2  x d x = ∫ ( 1 cos 2  x + sin  x cos 2  x ) d x = tan  x + ∫ sin  x cos 2  x d x {\displaystyle {\begin{aligned}I_{25}&=\int {\frac {1}{1-\sin x}}dx\\&=\int {\frac {1+\sin x}{1-\sin ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}+{\frac {\sin x}{\cos ^{2}x}}\right)dx\\&=\tan x+\int {\frac {\sin x}{\cos ^{2}x}}dx\end{aligned}}}
t = cos  x {\displaystyle t=\cos x} とおくと d t = − sin  x d x {\displaystyle dt=-\sin xdx} なので、
I 25 = tan  x − ∫ d t t 2 = tan  x + 1 t + C = tan  x + 1 cos  x + C {\displaystyle {\begin{aligned}I_{25}&=\tan x-\int {\frac {dt}{t^{2}}}\\&=\tan x+{\frac {1}{t}}+C\\&=\tan x+{\frac {1}{\cos x}}+C\end{aligned}}}
I 26 = ∫ x x 3 4 + 1 d x {\displaystyle {\begin{aligned}I_{26}&=\int {\frac {\sqrt {x}}{{\sqrt[{4}]{x^{3}}}+1}}dx\end{aligned}}}
t = x 3 4 {\displaystyle t={\sqrt[{4}]{x^{3}}}} とおくと d t = 3 4 x 4 d x {\displaystyle dt={\frac {3}{4{\sqrt[{4}]{x}}}}dx} なので、
I 26 = 4 3 ∫ t t + 1 d t = 4 3 ∫ ( 1 − 1 t + 1 ) d t = 4 3 t − 4 3 log  | t + 1 | + C = 4 3 x 3 4 − 4 3 log  ( x 3 4 + 1 ) + C {\displaystyle {\begin{aligned}I_{26}&={\frac {4}{3}}\int {\frac {t}{t+1}}dt\\&={\frac {4}{3}}\int \left(1-{\frac {1}{t+1}}\right)dt\\&={\frac {4}{3}}t-{\frac {4}{3}}\log |t+1|+C\\&={\frac {4}{3}}{\sqrt[{4}]{x^{3}}}-{\frac {4}{3}}\log \left({\sqrt[{4}]{x^{3}}}+1\right)+C\end{aligned}}}
I 27 = ∫ log  ( x 2 − 1 ) d x = ∫ log  ( x + 1 ) ( x − 1 ) d x = ∫ ( log  | x + 1 | + log  | x − 1 | ) d x = ( x + 1 ) log  | x + 1 | − x + ( x − 1 ) log  | x − 1 | − x + C = ( x + 1 ) log  | x + 1 | + ( x − 1 ) log  | x − 1 | − 2 x + C {\displaystyle {\begin{aligned}I_{27}&=\int \log(x^{2}-1)dx\\&=\int \log(x+1)(x-1)dx\\&=\int \left(\log |x+1|+\log |x-1|\right)dx\\&=(x+1)\log |x+1|-x+(x-1)\log |x-1|-x+C\\&=(x+1)\log |x+1|+(x-1)\log |x-1|-2x+C\end{aligned}}}
I 28 = ∫ e 3 x e x − 1 d x {\displaystyle {\begin{aligned}I_{28}&=\int {\frac {e^{3x}}{e^{x}-1}}dx\end{aligned}}}
t = e x {\displaystyle t=e^{x}} とおくと d t = e x d x {\displaystyle dt=e^{x}dx} なので、
I 28 = ∫ t 2 t − 1 d t = ∫ ( t + 1 + 1 t − 1 ) d t = 1 2 t 2 + t + log  | t − 1 | + C = 1 2 e 2 x + e x + log  | e x − 1 | + C {\displaystyle {\begin{aligned}I_{28}&=\int {\frac {t^{2}}{t-1}}dt\\&=\int \left(t+1+{\frac {1}{t-1}}\right)dt\\&={\frac {1}{2}}t^{2}+t+\log |t-1|+C\\&={\frac {1}{2}}e^{2x}+e^{x}+\log |e^{x}-1|+C\end{aligned}}}
I 29 = ∫ e x e x − e − x d x {\displaystyle {\begin{aligned}I_{29}&=\int {\frac {e^{x}}{e^{x}-e^{-x}}}dx\\\end{aligned}}}
分子分母に e x {\displaystyle e^{x}} を掛けて、
I 29 = ∫ e 2 x e 2 x − 1 d x {\displaystyle {\begin{aligned}I_{29}&=\int {\frac {e^{2x}}{e^{2x}-1}}dx\end{aligned}}}
t = e 2 x {\displaystyle t=e^{2x}} とおくと d t = 2 e 2 x d x {\displaystyle dt=2e^{2x}dx} なので、
I 29 = 1 2 ∫ d t t − 1 = 1 2 log  | t − 1 | + C = 1 2 log  | e 2 x − 1 | + C {\displaystyle {\begin{aligned}I_{29}&={\frac {1}{2}}\int {\frac {dt}{t-1}}\\&={\frac {1}{2}}\log |t-1|+C\\&={\frac {1}{2}}\log |e^{2x}-1|+C\end{aligned}}}
I 30 = ∫ d x 3 x 2 − 4 x − 2 = ∫ 3 ( 3 x − 2 − 10 ) ( 3 x − 2 + 10 ) d x = ∫ 3 2 10 ( 1 3 x − 2 − 10 − 1 3 x − 2 + 10 ) d x = 1 2 10 ( log  | 3 x − 2 − 10 | − log  | 3 x − 2 + 10 | ) + C = 1 2 10 log  | 3 x − 2 − 10 3 x − 2 + 10 | + C {\displaystyle {\begin{aligned}I_{30}&=\int {\frac {dx}{3x^{2}-4x-2}}\\&=\int {\frac {3}{\left(3x-2-{\sqrt {10}}\right)\left(3x-2+{\sqrt {10}}\right)}}dx\\&=\int {\frac {3}{2{\sqrt {10}}}}\left({\frac {1}{3x-2-{\sqrt {10}}}}-{\frac {1}{3x-2+{\sqrt {10}}}}\right)dx\\&={\frac {1}{2{\sqrt {10}}}}\left(\log |3x-2-{\sqrt {10}}|-\log |3x-2+{\sqrt {10}}|\right)+C\\&={\frac {1}{2{\sqrt {10}}}}\log \left|{\frac {3x-2-{\sqrt {10}}}{3x-2+{\sqrt {10}}}}\right|+C\end{aligned}}}
I 31 = ∫ d x 2 x 2 − 4 x + 3 = 1 2 ∫ d x ( x − 1 ) 2 + 1 2 {\displaystyle {\begin{aligned}I_{31}&=\int {\frac {dx}{\sqrt {2x^{2}-4x+3}}}\\&={\frac {1}{\sqrt {2}}}\int {\frac {dx}{\sqrt {(x-1)^{2}+{\frac {1}{2}}}}}\\\end{aligned}}}
t = x − 1 + ( x − 1 ) 2 + 1 2 {\displaystyle t=x-1+{\sqrt {(x-1)^{2}+{\frac {1}{2}}}}} とおくと d t = x − 1 + ( x − 1 ) 2 + 1 2 ( x − 1 ) 2 + 1 2 d x {\displaystyle dt={\frac {x-1+{\sqrt {(x-1)^{2}+{\frac {1}{2}}}}}{\sqrt {(x-1)^{2}+{\frac {1}{2}}}}}dx} なので、
I 31 = 1 2 ∫ d t t = 1 2 log  | t | + C = 1 2 log  ( x − 1 + ( x − 1 ) 2 + 1 2 ) + C = 1 2 log  ( x − 1 + x 2 − 2 x + 3 2 ) + C {\displaystyle {\begin{aligned}I_{31}&={\frac {1}{\sqrt {2}}}\int {\frac {dt}{t}}\\&={\frac {1}{\sqrt {2}}}\log |t|+C\\&={\frac {1}{\sqrt {2}}}\log \left(x-1+{\sqrt {(x-1)^{2}+{\frac {1}{2}}}}\right)+C\\&={\frac {1}{\sqrt {2}}}\log \left(x-1+{\sqrt {x^{2}-2x+{\frac {3}{2}}}}\right)+C\end{aligned}}}
I 32 = ∫ tan  x cos 3  x d x = ∫ sin  x cos 4  x d x {\displaystyle {\begin{aligned}I_{32}&=\int {\frac {\tan x}{\cos ^{3}x}}dx\\&=\int {\frac {\sin x}{\cos ^{4}x}}dx\\\end{aligned}}}
t = cos  x {\displaystyle t=\cos x} とおくと d t = − sin  x d x {\displaystyle dt=-\sin xdx} なので、
I 32 = − ∫ d t t 4 = 1 3 t 3 + C = 1 3 cos 3  x + C {\displaystyle {\begin{aligned}I_{32}&=-\int {\frac {dt}{t^{4}}}\\&={\frac {1}{3t^{3}}}+C\\&={\frac {1}{3\cos ^{3}x}}+C\end{aligned}}}
I 33 = ∫ d x 1 + cos  x = ∫ 1 − cos  x 1 − cos 2  x d x = ∫ ( 1 sin 2  x − cos  x sin 2  x ) d x = − 1 tan  x − ∫ cos  x sin 2  x d x {\displaystyle {\begin{aligned}I_{33}&=\int {\frac {dx}{1+\cos x}}\\&=\int {\frac {1-\cos x}{1-\cos ^{2}x}}dx\\&=\int \left({\frac {1}{\sin ^{2}x}}-{\frac {\cos x}{\sin ^{2}x}}\right)dx\\&=-{\frac {1}{\tan x}}-\int {\frac {\cos x}{\sin ^{2}x}}dx\end{aligned}}}
t = sin  x {\displaystyle t=\sin x} とおくと d t = cos  x d x {\displaystyle dt=\cos xdx} なので、
I 33 = − 1 tan  x − ∫ d t t 2 = − 1 tan  x + 1 t + C = − 1 tan  x + 1 sin  x + C {\displaystyle {\begin{aligned}I_{33}&=-{\frac {1}{\tan x}}-\int {\frac {dt}{t^{2}}}\\&=-{\frac {1}{\tan x}}+{\frac {1}{t}}+C\\&=-{\frac {1}{\tan x}}+{\frac {1}{\sin x}}+C\end{aligned}}}
別解
I 33 = ∫ d x 1 + cos  x = ∫ d x 2 cos 2  x 2 = tan  x 2 + C {\displaystyle {\begin{aligned}I_{33}&=\int {\frac {dx}{1+\cos x}}\\&=\int {\frac {dx}{2\cos ^{2}{\frac {x}{2}}}}\\&=\tan {\frac {x}{2}}+C\end{aligned}}}
I 34 = ∫ x tan 2  x d x = ∫ ( x cos 2  x − x ) d x = x tan  x − ∫ tan  x d x − 1 2 x 2 = x tan  x + log  | cos  x | − 1 2 x 2 + C {\displaystyle {\begin{aligned}I_{34}&=\int x\tan ^{2}xdx\\&=\int \left({\frac {x}{\cos ^{2}x}}-x\right)dx\\&=x\tan x-\int \tan xdx-{\frac {1}{2}}x^{2}\\&=x\tan x+\log |\cos x|-{\frac {1}{2}}x^{2}+C\end{aligned}}}
I 35 = ∫ e x cos  x d x = e x cos  x + ∫ e x sin  x d x = e x cos  x + e x sin  x − ∫ e x cos  x d x = e x cos  x + e x sin  x − I 35 {\displaystyle {\begin{aligned}I_{35}&=\int e^{x}\cos xdx\\&=e^{x}\cos x+\int e^{x}\sin xdx\\&=e^{x}\cos x+e^{x}\sin x-\int e^{x}\cos xdx\\&=e^{x}\cos x+e^{x}\sin x-I_{35}\end{aligned}}}
なので、
2 I 35 = e x sin  x + e x cos  x + C ′ I 35 = e x 2 ( sin  x + cos  x ) + C {\displaystyle {\begin{aligned}2I_{35}&=e^{x}\sin x+e^{x}\cos x+C'\\I_{35}&={\frac {e^{x}}{2}}(\sin x+\cos x)+C\end{aligned}}}
I 36 = ∫ x 2 − x − 1 d x = ∫ ( x − 1 2 ) 2 − 5 4 d x {\displaystyle {\begin{aligned}I_{36}&=\int {\sqrt {x^{2}-x-1}}dx\\&=\int {\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}dx\\\end{aligned}}}
t = x − 1 2 + ( x − 1 2 ) 2 − 5 4 {\displaystyle t=x-{\frac {1}{2}}+{\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}} とすると d t = x − 1 2 + ( x − 1 2 ) 2 − 5 4 ( x − 1 2 ) 2 − 5 4 d x {\displaystyle dt={\frac {x-{\frac {1}{2}}+{\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}}{\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}}dx} なので、
I 36 = ∫ ( t 2 + 5 8 t ) 2 − 5 4 t d t = 1 2 ∫ ( t 2 − 5 4 t + 25 32 t 3 ) d t = 1 2 ( t 2 4 − 5 4 log  | t | − 25 64 t 2 ) + C = 1 2 ( ( x − 1 2 ) x 2 − x − 1 − 5 4 log  | x − 1 2 + x 2 − x − 1 | ) + C {\displaystyle {\begin{aligned}I_{36}&=\int {\frac {\left({\frac {t}{2}}+{\frac {5}{8t}}\right)^{2}-{\frac {5}{4}}}{t}}dt\\&={\frac {1}{2}}\int \left({\frac {t}{2}}-{\frac {5}{4t}}+{\frac {25}{32t^{3}}}\right)dt\\&={\frac {1}{2}}\left({\frac {t^{2}}{4}}-{\frac {5}{4}}\log |t|-{\frac {25}{64t^{2}}}\right)+C\\&={\frac {1}{2}}\left(\left(x-{\frac {1}{2}}\right){\sqrt {x^{2}-x-1}}-{\frac {5}{4}}\log \left|x-{\frac {1}{2}}+{\sqrt {x^{2}-x-1}}\right|\right)+C\end{aligned}}}
別解
t = x − 1 2 + ( x − 1 2 ) 2 − 5 4 {\displaystyle t=x-{\frac {1}{2}}+{\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}} と置く。...1
{ t − ( x − 1 2 ) } 2 = ( ( x − 1 2 ) 2 − 5 4 ) 2 {\displaystyle \left\{t-\left(x-{\frac {1}{2}}\right)\right\}^{2}=\left({\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}\right)^{2}}
2 t ( x − 1 2 ) = t 2 + 5 4 {\displaystyle 2t\left(x-{\frac {1}{2}}\right)=t^{2}+{\frac {5}{4}}}
x − 1 2 = 1 2 ( t + 5 4 t ) {\displaystyle x-{\frac {1}{2}}={\frac {1}{2}}\left(t+{\frac {5}{4t}}\right)} ...2
12より、 ( x − 1 2 ) 2 − 5 4 = t − ( x − 1 2 ) = t − 1 2 ( t + 5 4 t ) = 1 2 ( t − 5 4 t ) {\displaystyle {\sqrt {\left(x-{\frac {1}{2}}\right)^{2}-{\frac {5}{4}}}}=t-\left(x-{\frac {1}{2}}\right)=t-{\frac {1}{2}}\left(t+{\frac {5}{4t}}\right)={\frac {1}{2}}\left(t-{\frac {5}{4t}}\right)} ...3
2をtで微分すると、 d x d t = 1 2 − 5 8 t 2 = 1 2 ( 1 − 5 4 t 2 ) {\displaystyle {\frac {dx}{dt}}={\frac {1}{2}}-{\frac {5}{8t^{2}}}={\frac {1}{2}}\left(1-{\frac {5}{4t^{2}}}\right)}
d x = 1 2 ( 1 − 5 4 t 2 ) d t {\displaystyle dx={\frac {1}{2}}\left(1-{\frac {5}{4t^{2}}}\right)dt} ...4
234より
I 36 = ∫ 1 2 ( t − 5 4 t ) ⋅ 1 2 ( 1 − 5 4 t 2 ) d t = 1 4 ∫ ( t − 10 4 t + 25 16 t 3 ) d t = 1 4 ( t 2 2 − 5 2 log  | t | − 25 32 t 2 ) + C = 1 8 ( t 2 − 25 16 t 2 ) − 5 8 log  | t | + C = 1 2 ⋅ 1 2 ( t + 5 4 t ) 1 2 ( t − 5 4 t ) − 5 8 log  | t | + C = 1 2 ( x − 1 2 ) x 2 − x − 1 − 5 8 log  | x − 1 2 + x 2 − x − 1 | + C {\displaystyle {\begin{aligned}I_{36}&=\int {\frac {1}{2}}\left(t-{\frac {5}{4t}}\right)\cdot {\frac {1}{2}}\left(1-{\frac {5}{4t^{2}}}\right)dt\\&={\frac {1}{4}}\int \left(t-{\frac {10}{4t}}+{\frac {25}{16t^{3}}}\right)dt\\&={\frac {1}{4}}\left({\frac {t^{2}}{2}}-{\frac {5}{2}}\log \left|t\right|-{\frac {25}{32t^{2}}}\right)+C\\&={\frac {1}{8}}\left(t^{2}-{\frac {25}{16t^{2}}}\right)-{\frac {5}{8}}\log \left|t\right|+C\\&={\frac {1}{2}}\cdot {\frac {1}{2}}\left(t+{\frac {5}{4t}}\right){\frac {1}{2}}\left(t-{\frac {5}{4t}}\right)-{\frac {5}{8}}\log \left|t\right|+C\\&={\frac {1}{2}}\left(x-{\frac {1}{2}}\right){\sqrt {x^{2}-x-1}}-{\frac {5}{8}}\log \left|x-{\frac {1}{2}}+{\sqrt {x^{2}-x-1}}\right|+C\\\end{aligned}}}
I 37 = ∫ sin  2 x ⋅ cos  3 x d x = 1 2 ∫ ( sin  5 x − sin  x ) d x = − 1 10 cos  5 x + 1 2 cos  x + C {\displaystyle {\begin{aligned}I_{37}&=\int \sin 2x\cdot \cos 3xdx\\&={\frac {1}{2}}\int (\sin 5x-\sin x)dx\\&=-{\frac {1}{10}}\cos 5x+{\frac {1}{2}}\cos x+C\end{aligned}}}
I 38 = ∫ ( sin  x + 1 sin  x ) 2 d x = ∫ ( sin 2  x + 2 + 1 sin 2  x ) d x = ∫ ( 1 − cos  2 x 2 + 2 + 1 sin 2  x ) d x = 5 2 x − sin  2 x 4 − 1 tan  x + C {\displaystyle {\begin{aligned}I_{38}&=\int \left(\sin x+{\frac {1}{\sin x}}\right)^{2}dx\\&=\int \left(\sin ^{2}x+2+{\frac {1}{\sin ^{2}x}}\right)dx\\&=\int \left({\frac {1-\cos 2x}{2}}+2+{\frac {1}{\sin ^{2}x}}\right)dx\\&={\frac {5}{2}}x-{\frac {\sin 2x}{4}}-{\frac {1}{\tan x}}+C\end{aligned}}}
I 39 = ∫ 1 + cos 3  x cos 2  x d x = ∫ ( 1 cos 2  x + cos  x ) d x = tan  x + sin  x + C {\displaystyle {\begin{aligned}I_{39}&=\int {\frac {1+\cos ^{3}x}{\cos ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}+\cos x\right)dx\\&=\tan x+\sin x+C\end{aligned}}}
I 40 = ∫ d x 1 + sin  x = ∫ 1 − sin  x 1 − sin 2  x d x = ∫ ( 1 cos 2  x − sin  x cos 2  x ) d x = tan  x − ∫ sin  x cos 2  x d x {\displaystyle {\begin{aligned}I_{40}&=\int {\frac {dx}{1+\sin x}}\\&=\int {\frac {1-\sin x}{1-\sin ^{2}x}}dx\\&=\int \left({\frac {1}{\cos ^{2}x}}-{\frac {\sin x}{\cos ^{2}x}}\right)dx\\&=\tan x-\int {\frac {\sin x}{\cos ^{2}x}}dx\\\end{aligned}}}
t = cos  x {\displaystyle t=\cos x} とおくと d t = − sin  x d x {\displaystyle dt=-\sin xdx} なので、
I 40 = tan  x + ∫ d t t 2 = tan  x − 1 t + C = tan  x − 1 cos  x + C {\displaystyle {\begin{aligned}I_{40}&=\tan x+\int {\frac {dt}{t^{2}}}\\&=\tan x-{\frac {1}{t}}+C\\&=\tan x-{\frac {1}{\cos x}}+C\end{aligned}}}
I 41 = ∫ d x e x + 1 {\displaystyle {\begin{aligned}I_{41}&=\int {\frac {dx}{e^{x}+1}}\end{aligned}}}
t = e x {\displaystyle t=e^{x}} とおくと d t = e x d x {\displaystyle dt=e^{x}dx} なので、
I 41 = ∫ d t t ( t + 1 ) = ∫ ( 1 t − 1 t + 1 ) d t = log  | t | − log  | t + 1 | + C = x − log  ( e x + 1 ) + C {\displaystyle {\begin{aligned}I_{41}&=\int {\frac {dt}{t(t+1)}}\\&=\int \left({\frac {1}{t}}-{\frac {1}{t+1}}\right)dt\\&=\log |t|-\log |t+1|+C\\&=x-\log(e^{x}+1)+C\end{aligned}}}
I 42 = ∫ ( log  x ) 2 d x = x ( log  x ) 2 − 2 ∫ log  x d x = x ( log  x ) 2 − 2 x log  x + 2 x + C {\displaystyle {\begin{aligned}I_{42}&=\int \left(\log x\right)^{2}dx\\&=x\left(\log x\right)^{2}-2\int \log xdx\\&=x\left(\log x\right)^{2}-2x\log x+2x+C\end{aligned}}}
I 43 = ∫ 1 − x 2 x d x {\displaystyle {\begin{aligned}I_{43}&=\int {\frac {\sqrt {1-x^{2}}}{x}}dx\end{aligned}}}
t = 1 − x 2 {\displaystyle t={\sqrt {1-x^{2}}}} とおくと d t = − x 1 − x 2 d x {\displaystyle dt=-{\frac {x}{\sqrt {1-x^{2}}}}dx} なので、
I 43 = ∫ t 2 t 2 − 1 d t = ∫ ( 1 + 1 t 2 − 1 ) d t = ∫ ( 1 + 1 2 ( t − 1 ) − 1 2 ( t + 1 ) ) d t = t + log  | t − 1 t + 1 | + C = 1 − x 2 + log  1 − 1 − x 2 | x | + C {\displaystyle {\begin{aligned}I_{43}&=\int {\frac {t^{2}}{t^{2}-1}}dt\\&=\int \left(1+{\frac {1}{t^{2}-1}}\right)dt\\&=\int \left(1+{\frac {1}{2(t-1)}}-{\frac {1}{2(t+1)}}\right)dt\\&=t+\log {\sqrt {\left|{\frac {t-1}{t+1}}\right|}}+C\\&={\sqrt {1-x^{2}}}+\log {\frac {1-{\sqrt {1-x^{2}}}}{|x|}}+C\end{aligned}}}
I 44 = ∫ d x x 2 x − x 2 {\displaystyle {\begin{aligned}I_{44}&=\int {\frac {dx}{x{\sqrt {2x-x^{2}}}}}\\\end{aligned}}}
t = 1 x {\displaystyle t={\frac {1}{x}}} とおくと d t = − 1 x 2 d x {\displaystyle dt=-{\frac {1}{x^{2}}}dx} なので、
I 44 = − ∫ d t 2 t − 1 = − 2 t − 1 + C = − 2 − x x + C {\displaystyle {\begin{aligned}I_{44}&=-\int {\frac {dt}{\sqrt {2t-1}}}\\&=-{\sqrt {2t-1}}+C\\&=-{\sqrt {\frac {2-x}{x}}}+C\end{aligned}}}