# Calculus/Indefinite integral/Solutions

$\int {\frac {3x}{2}}dx$
$F$
    $F'(x)={\frac {3x}{2}}$
We know that
    ${\frac {d}{dx}}x^{2}=2x$
So we need to find a constant, $a$ , such that
    ${\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}$
Solving for $a$ , we get
    $2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}$
So
    $\int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C$
Check your answer by taking the derivative of the function you've found and checking that it matches the integrand:
    ${\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}$
$F$
    $F'(x)={\frac {3x}{2}}$
We know that
    ${\frac {d}{dx}}x^{2}=2x$
So we need to find a constant, $a$ , such that
    ${\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}$
Solving for $a$ , we get
    $2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}$
So
    $\int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C$
Check your answer by taking the derivative of the function you've found and checking that it matches the integrand:
    ${\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}$
$f(x)=2x^{4}$
${\frac {d}{dx}}x^{5}=5x^{4}$ We need to find a constant, $a$, such that ${\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}$ Solving for $a$, we get $5ax^{4}=2x^{4}\implies a={\frac {2}{5}}$ So the general antiderivative will be $\mathbf {{\frac {2}{5}}x^{5}+C}$ Check your answer by taking the derivative of the antiderivative you've found and checking that you get back the function you started with:
${\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}$
${\frac {d}{dx}}x^{5}=5x^{4}$ We need to find a constant, $a$, such that ${\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}$ Solving for $a$, we get $5ax^{4}=2x^{4}\implies a={\frac {2}{5}}$ So the general antiderivative will be $\mathbf {{\frac {2}{5}}x^{5}+C}$ Check your answer by taking the derivative of the antiderivative you've found and checking that you get back the function you started with:
${\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}$
$\int (7x^{2}+3\cos(x)-e^{x})dx$
${\begin{aligned}\int (7x^{2}+3\cos(x)-e^{x})dx&=7\int x^{2}dx+3\int \cos(x)dx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin(x)-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin(x)-e^{x}+C} \end{aligned}}$
${\begin{aligned}\int (7x^{2}+3\cos(x)-e^{x})dx&=7\int x^{2}dx+3\int \cos(x)dx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin(x)-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin(x)-e^{x}+C} \end{aligned}}$
$\int ({\frac {2}{5x}}+\sin(x))dx$
${\begin{aligned}\int ({\frac {2}{5x}}+\sin(x))dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin(x)dx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos(x)+C} \end{aligned}}$
${\begin{aligned}\int ({\frac {2}{5x}}+\sin(x))dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin(x)dx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos(x)+C} \end{aligned}}$
$\int x\sin(2x^{2})dx$
$u=2x^{2}$
$u=2x^{2}$
$du=4xdx$
$dx={\frac {du}{4x}}$
${\begin{aligned}\int x\sin(2x^{2})dx&=\int x\sin(u){\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin(u)du\\&=-{\frac {\cos(u)}{4}}+C\\&=-\mathbf {{\frac {\cos(2x^{2})}{4}}+C} \end{aligned}}$
$u=2x^{2}$
$du=4xdx$
$dx={\frac {du}{4x}}$
${\begin{aligned}\int x\sin(2x^{2})dx&=\int x\sin(u){\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin(u)du\\&=-{\frac {\cos(u)}{4}}+C\\&=-\mathbf {{\frac {\cos(2x^{2})}{4}}+C} \end{aligned}}$
$\int -3\cos(x)e^{\sin(x)}dx$
$u=\sin(x)$
$du=\cos(x)dx$
$dx={\frac {du}{\cos(x)}}$
${\begin{aligned}\int -3\cos(x)e^{\sin(x)}dx&=-3\int \cos(x)e^{u}{\frac {du}{\cos(x)}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin(x)}+C} \end{aligned}}$
$u=\sin(x)$
$du=\cos(x)dx$
$dx={\frac {du}{\cos(x)}}$
${\begin{aligned}\int -3\cos(x)e^{\sin(x)}dx&=-3\int \cos(x)e^{u}{\frac {du}{\cos(x)}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin(x)}+C} \end{aligned}}$
$\int {\frac {2x-5}{x^{3}}}dx$
$u=2x-5$
$dv={\frac {dx}{x^{3}}}$
$du=2dx$
$v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}$
${\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}$
$du=2dx$
$v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}$
${\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}$
$\int (2x-1)e^{-3x+1}dx$
$u=2x-1$
$dv=e^{-3x+1}dx$
Then $du=2dx$ and $v=\int e^{-3x+1}dx$ To evaluate $v$, make the substitution $w=-3x+1$; $dw=-3dx$; $dx={\frac {-dw}{3}}$. Then $v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}$. So
${\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}$
$u=2x-1$
$dv=e^{-3x+1}dx$
Then $du=2dx$ and $v=\int e^{-3x+1}dx$ To evaluate $v$, make the substitution $w=-3x+1$; $dw=-3dx$; $dx={\frac {-dw}{3}}$. Then $v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}$. So
${\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}$
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