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

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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


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