#### Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 11

$\frac{1}{6} \log \left|\frac{x^{6}}{x^{6}+1}\right|+C$

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

$\int \frac{1}{x\left(x^{6}+1\right)} d x$

Solution:

$Let\: \: I=\int \frac{1}{x\left(x^{6}+1\right)} d x$

\begin{aligned} &=\int \frac{x^{5}}{x^{5}\left\{x\left(x^{6}+1\right)\right\}} d x \\ &=\int \frac{x^{5}}{x^{6}\left(x^{6}+1\right)} d x \end{aligned}

$Put\: \: x^{6}=t\Rightarrow 6x^{5}dx=dt\Rightarrow x^{5}dx=\frac{dt}{6}$

$Then\: \: I=\int \frac{1}{t\left(t+1\right)} \frac{dt}{6}$

\begin{aligned} &=\frac{1}{6} \int \frac{1}{t^{2}+t} d t \\ &=\frac{1}{6} \int \frac{1}{t^{2}+2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \\ &=\frac{1}{6} \int \frac{1}{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}} d t \end{aligned}

$Put\: \: t+\frac{1}{2}=u\Rightarrow dt=du$

\begin{aligned} =\frac{1}{6} \int \frac{1}{u^{2}-(\frac{1}{2})^{2}} d t \end{aligned}\begin{aligned} =\frac{1}{6} \int \frac{1}{u^{2}-(\frac{1}{2})^{2}} d u \end{aligned}

\begin{aligned} &=\frac{1}{6} \cdot \frac{1}{2 \times\left(\frac{1}{2}\right)} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{1}{6} \log \left|\frac{\frac{2u-1}{2}}{\frac{2u+1}{2}}\right|+C \\ &=\frac{1}{6} \log \left|\frac{2 u-1}{2 u+1}\right|+C \end{aligned}

\begin{aligned} &=\frac{1}{6} \log \left|\frac{2\left(t+\frac{1}{2}\right)-1}{2\left(t+\frac{1}{2}\right)+1}\right|+C \quad\left[\because t+\frac{1}{2}=u\right] \\ &=\frac{1}{6} \log \left|\frac{2 t+1-1}{2 t+1+1}\right|+C \\ &=\frac{1}{6} \log \left|\frac{2 t}{2 t+2}\right|+C \\ &=\frac{1}{6} \log \left|\frac{t}{t+1}\right|+C \\ &=\frac{1}{6} \log \left|\frac{x^{6}}{x^{6}+1}\right|+C \quad\left[\because t=x^{6}\right] \end{aligned}