#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 10 Maths Textbook Solution.

$\frac{1}{2 \sqrt{5}}\left(\frac{x^{2}-1}{\sqrt{5} x}\right)-\frac{1}{2}\left(\frac{x^{2}+1}{x}\right)+c$

Given:

$\int \frac{1}{x^{4}+3 x^{2}+1} d x$

Hint:

The given integral function can be converted into

$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$

Solution

\begin{aligned} I &=\int \frac{1}{x^{4}+3 x^{2}+1} d x \\ &=\int \frac{\frac{1}{x^{2}}}{x^{2}+3+\frac{1}{x^{2}}} d x \end{aligned}(Numerator and denominator divided by $x^{2}$)

\begin{aligned} &=\frac{1}{2} \int \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+3+\frac{1}{x^{2}}} d x \\ &=\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+5} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}+1} d x\right] \end{aligned}

\begin{aligned} \text { Assume } t &=x-\frac{1}{x} \text { and } z=x+\frac{1}{x} \\ d t &=\left(1+\frac{1}{x^{2}}\right) d x \text { and } d z=\left(1-\frac{1}{x^{2}}\right) d x \\ &=\frac{1}{2}\left[\int \frac{d t}{t^{2}+5}-\int \frac{d z}{z^{2}+1}\right] \end{aligned}

Using identity $\int \frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\frac{x}{a}+c$

$I=\frac{1}{2\sqrt{5}}\tan-\left ( \frac{t}{\sqrt{5}} \right )-\frac{1}{2}\tan^{-1}z$

Substituting  $t \: as\: x-\frac{1}{x}and\: z\: as\: x+\frac{1}{x}$

\begin{aligned} &I=\frac{1}{2 \sqrt{5}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{5}}\right)-\frac{1}{2} \tan ^{-1}\left(x+\frac{1}{x}\right)+c \\ &\frac{1}{2 \sqrt{5}}\left(\frac{x^{2}-1}{\sqrt{5} x}\right)-\frac{1}{2}\left(\frac{x^{2}+1}{x}\right)+c \end{aligned}

Where c is integrating constant.