#### Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 125 Maths Textbook Solution.

Answer: $\frac{1}{3}\left(\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{1}{\sqrt{5}} \tan ^{-1} \frac{x}{\sqrt{5}}\right)+C$

Given: $\int \frac{1}{\left(x^{2}+2\right)\left(x^{2}+5\right)} d x$

Hint: Using $\int \frac{1}{1+x^{2}} d x$

Explanation: let $I=\int \frac{1}{\left(x^{2}+2\right)\left(x^{2}+5\right)} d x$

$=\frac{1}{3} \int \frac{\left(x^{2}+5\right)-\left(x^{2}+2\right)}{\left(x^{2}+5\right)\left(x^{2}+2\right)} d x$

$=\frac{1}{3} \int\left(\frac{1}{x^{2}+2}-\frac{1}{x^{2}+5}\right) d x$

$=\frac{1}{3} \int \frac{1}{x^{2}+2} d x-\frac{1}{3} \int \frac{1}{x^{2}+5} d x$

$=\frac{1}{3} \int \frac{1}{x^{2}+(\sqrt{2})^{2}} d x-\frac{1}{3} \int \frac{1}{x^{2}+(\sqrt{5})^{2}} d x$

$=\frac{1}{3 \sqrt{2}} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)-\frac{1}{3 \sqrt{5}} \tan ^{-1}\left(\frac{x}{\sqrt{5}}\right)+C$

$=\frac{1}{3}\left(\frac{1}{\sqrt{2}} \tan ^{-1} \frac{x}{\sqrt{2}}-\frac{1}{\sqrt{5}} \tan ^{-1} \frac{x}{\sqrt{5}}\right)+C$