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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.14 Question 3

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Answer:  \frac{1}{a b} \tan ^{-1}\left|\frac{a x}{b}\right|+c

Hint: To solve this integral, use special integral formula.

Given:  \int \frac{1}{a^{2} x^{2}+b^{2}} d x

Solution: Let  I=\int \frac{1}{a^{2} x^{2}+b^{2}} d x=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\frac{b^{2}}{a^{2}}} d x

=\frac{1}{a^{2}} \int \frac{1}{x^{2}+\left(\frac{b}{a}\right)^{2}} d x

=\frac{1}{a^{2}} \cdot \frac{1}{\frac{b}{a}} \tan ^{-1}\left|\frac{x}{\frac{b}{a}}\right|+c \quad\quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left|\frac{x}{a}\right|+c\right]

=\frac{1}{a b} \tan ^{-1}\left|\frac{a x}{b}\right|+c

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