#### Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.16 question 7

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

Hint:

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

Given:

$\int \frac{x}{x^{4}+2x^{2}+3}dx$

Solution:

$Let\: \: \: I=\int \frac{x}{x^{4}+2x^{2}+3}dx$

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

$Then\: \: \: I=\int \frac{1}{t^{2}+2t+3}\frac{dt}{2}$

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

$Put\: \: u=t+1\Rightarrow du=dt$

Then\: \: \begin{aligned} I=\frac{1}{2} \int \frac{1}{u^{2}+(\sqrt{2})^{2}} d u \end{aligned}

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