Question

If $I=\int \frac{dx}{x^3\sqrt{x^2-1}}$, then $I$ equals

• A. $\frac{1}{2}\left(\frac{\sqrt{x^2-1}}{x^3}\right)+{\tan }^{-1}\sqrt{x^2-1}+c$
• B. $\frac{1}{2}\left(\frac{\sqrt{x^2-1}}{x^2}\right)+x{\tan }^{-1}\sqrt{x^2-1}+c$
• C. $\frac{1}{2}\left(\frac{\sqrt{x^2-1}}{x}\right)+{\tan }^{-1}\sqrt{x^2-1}+c$
• D. $\frac{1}{2}\left(\frac{\sqrt{x^2-1}}{x^2}\right)+{\tan }^{-1}\sqrt{x^2-1}+c$

Answer 1

The correct option is D $\frac{1}{2}\left(\frac{\sqrt{x^2-1}}{x^2}\right)+{\tan }^{-1}\sqrt{x^2-1}+c$

$I=\int \frac{dx}{x^4\sqrt{x^2-1}}$
Let $x^2-1=t^2\Rightarrow 2xdx=2tdt$
$\therefore I=\int \frac{t}{(t^2+1{)}^{2}t}dt=\int \frac{dt}{(t^2+1{)}^2}$
But
${\tan }^{-1}t=\int \frac{dt}{t^2+1}=\int 1.\frac{1}{t^2+1}dt$
$=\frac{t}{t^2+1}+\int t\frac{2t}{(t^2+1{)}^2}dt$
$=\frac{t}{t^2+1}+2\int \frac{t^2+1-1}{(t^2+1{)}^2}dt$
$=\frac{t}{t^2+1}+2{\tan }^{-1}t-2I$
$\Rightarrow I=\frac{1}{2}\frac{t}{t^2+1}+\frac{1}{2}{\tan }^{-1}t+c$
$\Rightarrow I=\frac{1}{2}(\frac{\sqrt{x^2-1}}{x^2}+{\tan }^{-1}\sqrt{x^2-1})+c$