_ ^ 1 ( x^2 + a^2 )( x^2 + b^2 ) dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{({x^2} + {a^2})({x^2} + {b^2})}}dx = } $

✓
$\frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$
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
$\frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) - \frac{1}{b}{\tan ^{ - 1}}\left( {\frac{x}{b}} \right) + c$

✓

Answer

Correct option: A.

$\frac{1}{{({a^2} - {b^2})}}\left[ {\frac{1}{b}{{\tan }^{ - 1}}\left( {\frac{x}{b}} \right) - \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} \right] + c$

a
(a)$\int_{}^{} {\frac{1}{{({x^2} + {b^2})({x^2} + {a^2})}}} \,dx$
$ = \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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