#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 8 Maths Textbook Solution.

Answer: The required value of the integral is,

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

Given:

$\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x$

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: Rewrite the given equation as.

$\int \frac{x^{2}+1}{x^{4}+7 x^{2}+1} d x$

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

\begin{aligned} I &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+7+2-2} dx \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}-2\right)+9} d x \\ &=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+9} d x \end{aligned}

$let(x-\frac{1}{x})=t\Rightarrow (1+\frac{1}{x^{2}})dx=dt$

Evaluating the obtained integration.

\begin{aligned} &I=\int \frac{d t}{t^{2}+3^{2}} \\ &I=\frac{1}{3} \tan -1\left(\frac{t}{3}\right)+c \\ &I=\frac{1}{3} \tan ^{-1}\left(\frac{\left.x-\frac{1}{x}\right)+c}{3}\right)+c \\ &I=\frac{1}{3} \tan ^{-1}\left(\frac{x^{2}-1}{3 x}\right)+c &\text { (Substituting } t \text { as }\left(x-\frac{1}{x}\right) \text { ) } \\ \end{aligned}