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

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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