#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 7 Maths Textbook Solution.

Answer: The required value of the integration is,

$\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$

Hint: Use the identity  formula $\int \frac{1}{x^{2}+a^{2}}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+c$

Given:

$I=\int \frac{x^{2}-1}{x^{4}+1} d x$

Solution: Given equation can be written as

$I=\frac{x^{2}-1}{x^{4}+1} d x$

Divide numerator and denominator by $x^{2}$

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

Let$x+\frac{1}{x}=t$

$\left ( 1-\frac{1}{x^{2}}\right )dx=dt$     differentiating on both sides with respect to t,

\begin{aligned} I &=\int \frac{d t}{t^{2}-2} \\ &=\int \frac{d t}{(t)^{2}-\left(\sqrt{2)^{2}}\right.} \\ &=\frac{1}{2 \sqrt{2}} \log \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+c \end{aligned}

Re- substitute$t=x+\frac{1}{x}$  into the above equation.

$\frac{1}{2 \sqrt{2}} \log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$

Where c is an integrating constant.