#### Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 12 maths

$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+C$

Hint:

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

Given:

$\int \frac{x}{x^{4}-x^{2}+1}dx$

Solution:

$Let\: \: I=\int \frac{x}{x^{4}-x^{2}+1}dx$

$=\int \frac{x}{(x^{2})^{2}-x^{2}+1}dx$

$Put\: \: x^{2}=t\Rightarrow 2x\, \, dx=dt\Rightarrow x\, dx=\frac{dt}{2}$

$=\int \frac{1}{t^{2}-t+1}\, \frac{dt}{2}$

\begin{aligned} &=\frac{1}{2} \int \frac{1}{t^{2}-2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\frac{1}{4}+1} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{1-4}{4}\right)} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{-3}{4}\right)} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}} d t \end{aligned}

$Let\: \: t-\frac{1}{2}=u\Rightarrow dt=du$

$Then \: \: I=\frac{1}{2} \int \frac{1}{u^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d u$

\begin{aligned} &=\frac{1}{2} \times \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \tan ^{-1}\left(\frac{u}{\frac{\sqrt{3}}{2}}\right)+C \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C \quad\left[\because u=t-\frac{1}{2}\right] \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\frac{2 t-1}{2}}{\frac{\sqrt{3}}{2}}\right)+C \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+C \quad\left[\because t=x^{2}\right] \end{aligned}