Integral Calculus Question 296
Question: $ \int_{{}}^{{}}{\frac{dx}{(1+x^{2})\sqrt{p^{2}+q^{2}{{({{\tan }^{-1}}x)}^{2}}}}}= $
Options:
A) $ \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
Show Answer
Answer:
Correct Answer: A
Solution:
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 [ q{{\tan }^{-1}}x+\sqrt{p^{2}+q^{2}{{({{\tan }^{-1}}x)}^{2}}} ]+c $ .
BETA
