#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 3 Maths Textbook Solution.

Answer: The required value of the integral is,

$I=\frac{1}{3 \sqrt{2}}\left(\frac{x-\frac{9}{x}}{3 \sqrt{2}}\right)+c$

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

Given:$\int \frac{\left(x^{2}+9\right)}{\left(x^{4}+81\right)} d x$

Solution: The equation can be written as

$I=\int \frac{1+\frac{9}{x^{2}}}{x^{2}+\frac{81}{x^{2}}} d x$

$=\int \frac{1+\frac{9}{x^{2}}}{\left(x-\frac{9}{x}\right)^{2}+18} d x$

Let, $x-\frac{9}{x}=t$ then,

Differentiating the above equation both sides.

$\left(1+\frac{9}{x^{2}}\right) d x=d t$ then the obtained equation become

$\int \frac{d t}{t^{2}+(3 \sqrt{2})^{2}}$

Now, using the identity rule,

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

Now, re-substituting $t=x-\frac{9}{x}$ then,

\begin{aligned} I &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x-\frac{9}{x}}{3 \sqrt{2}}\right)+c \\ &=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}-9}{3 \sqrt{2} x}\right)+c \end{aligned}

Where c is the integrating constant.