A) \[\frac{1}{({{a}^{2}}-{{b}^{2}})}\left[ \frac{1}{b}{{\tan }^{-1}}\left( \frac{x}{b} \right)-\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right) \right]+c\]
B) \[\frac{1}{({{b}^{2}}-{{a}^{2}})}\left[ \frac{1}{b}{{\tan }^{-1}}\left( \frac{x}{b} \right)-\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right) \right]+c\]
C) \[\frac{1}{b}{{\tan }^{-1}}\left( \frac{x}{b} \right)-\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right)+c\]
D) \[\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right)-\frac{1}{b}{{\tan }^{-1}}\left( \frac{x}{b} \right)+c\]
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{1}{({{x}^{2}}+{{b}^{2}})({{x}^{2}}+{{a}^{2}})}}\,dx\] \[=\frac{1}{{{a}^{2}}-{{b}^{2}}}\int_{{}}^{{}}{\left[ \frac{1}{{{x}^{2}}+{{b}^{2}}}-\frac{1}{{{x}^{2}}+{{a}^{2}}} \right]}\,dx\] \[=\frac{1}{({{a}^{2}}-{{b}^{2}})}\left[ \frac{1}{b}{{\tan }^{-1}}\left( \frac{x}{b} \right)-\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right) \right]+c\]. Note : Students should remember this question as a formula.
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