Question

$\int \frac{1}{(1+x^2)\sqrt{[p^2+q^2{\left({\tan }^{-1}x\right)}^2]}}dx.$

• A. $\frac{1}{q}\log [t-\sqrt{p^2+t^2}]$ where $t=q{\tan }^{-1}x.$
• B. $\frac{1}{t}\log [t+\sqrt{p^2+t^2}]$ where $t=q{\tan }^{-1}x.$
• C. $\frac{1}{q}\log [t+\sqrt{p^2+t^2}]$ where $t=q{\tan }^{-1}x.$
• D. $\frac{1}{q}\log [t+\sqrt{p^2-t^2}]$ where $t=q{\tan }^{-1}x.$

Answer 1

The correct option is C $\frac{1}{q}\log [t+\sqrt{p^2+t^2}]$ where $t=q{\tan }^{-1}x.$

Let $I=\int \frac{1}{(1+x^2)\sqrt{[p^2+q^2{\left({\tan }^{-1}x\right)}^2]}}dx$
Put $q{\tan }^{-1}x=t\Rightarrow q\frac{1}{1+x^2}dx=dt$
Therefore
$I=\frac{1}{q}\int \frac{dt}{\sqrt{(p^2+t^2)}}=\frac{1}{q}\log [t+\sqrt{p^2+t^2}]$
Where $t=q{\tan }^{-1}x$
Hence, option 'C' is correct.