_ ^ dx (1 + x^2 ) p^2 + q^2 ( ^ - 1 x) ^2 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} }}} = $

✓
$\frac{1}{q}\log [q{\tan ^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} ] + c$
B
$\log [q{\tan ^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} ] + c$
C
$\frac{2}{{3q}}{({p^2} + {q^2}{\tan ^{ - 1}}x)^{3/2}} + c$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{q}\log [q{\tan ^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} ] + c$

a
(a) Putting $q {\tan ^{ - 1}}x = t$

$ \Rightarrow \frac{q}{{1 + {x^2}}}dx = dt \Rightarrow \frac{1}{{1 + {x^2}}}dx = \frac{{dt}}{q}$
$ \Rightarrow \int_{}^{} {\frac{{dx}}{{(1 + {x^2})\sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} }}} = \frac{1}{q}\int_{}^{} {\frac{{dt}}{{\sqrt {{p^2} + {t^2}} }}} $
$ = \frac{1}{q}\log \left[ {q{{\tan }^{ - 1}}x + \sqrt {{p^2} + {q^2}{{({{\tan }^{ - 1}}x)}^2}} } \right] + c$.

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