Subject: Mathematics — Definite Integration
Question: Find the indefinite integral
$\int_{}^{}\ \left( \frac{1}{\sqrt[3]{x} + \sqrt[4]{4}} + \frac{ln(1 + \sqrt[6]{x}}{\sqrt[3]{x} + \sqrt{x}} \right)dx$
Answer: $\Rightarrow I = \frac{3}{2}x^{2/3} - \frac{12}{7}x^{7/2} + 2x^{1/2} - \frac{12}{5}x^{5/12} + 3x^{1/13} - 4x^{1/4}
$$${+ 6x^{1/6} - 12x^{1/12} + 12log\left| x^{1/12} + 1 \right|
}{+ 6\left\lbrack \left\{ \frac{\left( 1 + x^{1/6} \right)^{3}}{3} - \frac{3}{2}\left( 1 + x^{1/6} \right)^{2} + 3\left( 1 + x^{1/6} \right) \right\} ln\left( 1 + x^{1/6} \right) \right.\ 
}{- \left\{ \frac{\left( 1 + x^{1/6} \right)^{3}}{9} - \frac{3}{4}\left( 1 + x^{1/6} \right)^{2} + 3\left( 1 + x^{1/6} \right) \right\}\left. \ - \frac{\left\lbrack ln\left( 1 + x^{1/6} \right)^{2} \right\rbrack}{2} + c \right\rbrack}$$
Solution: $I = \int\left( \frac{1}{\sqrt[3]{x} + \sqrt[4]{x}} + \frac{ln(1 + \sqrt[6]{x})}{\sqrt[3]{x} + \sqrt{x}} \right)dx$\
$= \int\frac{1}{\sqrt[3]{x} + \sqrt[4]{x}}dx + \int\frac{ln(1 + \sqrt[6]{x})}{\sqrt[3]{x} + \sqrt{x}}dx$\
$I = I_{1} + I_{2}$ ......(i)\
where $I_{1} = \int_{}^{}\ \frac{1}{\sqrt[3]{x} + \sqrt[4]{x}}dx$ and
$I_{2} = \int_{}^{}\ \frac{ln(1 + \sqrt[6]{x})}{\sqrt[3]{x} + \sqrt{x}}dx$\
Let $x = y^{12}$ so that $dx = 12y^{11}dy$\
$${\therefore\ I_{1} = \int_{}^{}\ \frac{12y^{11}}{y^{4} + y^{3}}dy = 12\int_{}^{}\ \frac{y^{8}}{1 + y}dy
}{= 12\int_{}^{}\ \left( y^{7} - y^{6} + y^{5} - y^{4} + y^{3} - y^{2} + y - 1 + \frac{1}{y + 1} \right)dy
}{= 12\left\lbrack \frac{y^{8}}{8} - \frac{y^{7}}{7} + \frac{y^{6}}{6} - \frac{y^{5}}{5} + \frac{y^{4}}{4} - \frac{y^{3}}{3} \right.\ \left. \ + \frac{y^{2}}{2} - y + log|y + 1| \right\rbrack + c_{1}
}{= \frac{3}{2}x^{2/3} - \frac{12}{7}x^{7/12} + 2x^{1/2} - \frac{12}{5}x^{5/12} + 3x^{1/3}
}$$$- 4x^{1/4} + 6x^{1/6} - 12x^{1/12} + 12log\left| x^{1/12} + 1 \right| + c_{1}$
......(ii)\
Now,
$I_{2} = \int_{}^{}\ \frac{ln\left( 1 + (x)^{1/6} \right)}{(x)^{1/3} + (x)^{1/2}}dx$\
Let $x = z^{6}$ so that $dx = 6z^{5}dz$\
$$\therefore\ I_{2} = \int\frac{ln(1 + z)}{z^{2} + z^{3}},6z^{5}dz = \int\frac{6z^{3}ln(z + 1)}{z + 1}dz
$$Put $z + 1 = t$\
$${\Rightarrow dz = dt
}{\therefore\ I_{2} = \int\frac{6(t - 1)^{3}\ln t}{t}dt = 6\left( t^{2} - 3t + 3 - \frac{1}{t} \right)\ln\int tdt
}{= 6\left\lbrack \int\left( t^{2} - 3t + 3 \right)lntdt - \int\frac{1}{t}lntdt \right\rbrack
}{= 6\left\lbrack \left( \frac{t^{3}}{3} - \frac{3t^{2}}{2} + 3t \right)lnt - \int\left( \frac{t^{3}}{3} - \frac{3t^{2}}{2} + 3t \right) \cdot \frac{1}{t}dt \right.\ \left. \ \begin{matrix}
 - \frac{(lnt)^{2}}{2} \\
 - \frac{(lnt)^{2}}{2}
\end{matrix} \right\rbrack
}{= 6\left\lbrack \left( \frac{t^{3}}{3} - \frac{3t^{2}}{2} + 3t \right)lnt - \int\left( \frac{t^{2}}{3} - \frac{3}{2}t + 3 \right)dt \right.\ 
}{= 6\left\lbrack \left( \frac{t^{3}}{3} - \frac{3t^{2}}{2} + 3t \right)lnt - \left( \frac{t^{3}}{9} - \frac{3t^{2}}{4} + 3t \right) - \frac{(lnt)^{2}}{2} \right\rbrack + c_{2}
}$$Now, $t = 1 + z = 1 + x^{1/6}$\
$${\therefore I_{2} = 6\left\lbrack \left\{ \frac{\left( 1 + x^{1/6} \right)^{3}}{3} - \frac{3}{2}\left( 1 + x^{1/6} \right)^{2} + 3\left( 1 + x^{1/6} \right) \right\} \right.\ 
}{ln\left( 1 + x^{1/6} \right) - \left\{ \frac{\left( 1 + x^{1/6} \right)^{3}}{9} - \frac{3}{4}\left( 1 + x^{1/6} \right)^{2} \right.\ 
}$$$\left. \ \left. \ + 3\left( 1 + x^{1/6} \right) \right\} - \frac{\left\lbrack ln\left( 1 + x^{1/6} \right)^{2} \right\rbrack}{2} \right\rbrack + c_{2}$
......(iii)\
From (i), (ii) and (iii), we get $I = I_{1} + I_{2}$\
$${\Rightarrow I = \frac{3}{2}x^{2/3} - \frac{12}{7}x^{7/2} + 2x^{1/2} - \frac{12}{5}x^{5/12} + 3x^{1/13} - 4x^{1/4}
}{+ 6x^{1/6} - 12x^{1/12} + 12log\left| x^{1/12} + 1 \right|
}{+ 6\left\lbrack \left\{ \frac{\left( 1 + x^{1/6} \right)^{3}}{3} - \frac{3}{2}\left( 1 + x^{1/6} \right)^{2} + 3\left( 1 + x^{1/6} \right) \right\} ln\left( 1 + x^{1/6} \right) \right.\ 
}{- \left\{ \frac{\left( 1 + x^{1/6} \right)^{3}}{9} - \frac{3}{4}\left( 1 + x^{1/6} \right)^{2} + 3\left( 1 + x^{1/6} \right) \right\}\left. \ - \frac{\left\lbrack ln\left( 1 + x^{1/6} \right)^{2} \right\rbrack}{2} + c \right\rbrack}$$