Question

Evaluate: $\int \frac{1}{(x+1)\sqrt{x^2-1}}dx$

• A. $\frac{\sqrt{x+1}}{\sqrt{x-1}}+c$
• B. $\frac{\sqrt{x-1}}{\sqrt{x+1}}+c$
• C. $\frac{\sqrt{2x-1}}{\sqrt{x+1}}+c$
• D. None of these

Answer 1

Answer: (B)

$\frac{\sqrt{x-1}}{\sqrt{x+1}}+c$

$\int \frac{1}{(x-1)\sqrt{x^2-1}}dx$

Substitute $x=\sec u$

$dx=\sec u+\tan udu$

$=\frac{\sec u\tan u}{(\sec u+1)\sqrt{{\sec }^{2}y-1}}du$

${\sec }^{2}u-1={\tan }^{2}u$

$=\frac{\sec u}{\sec u+1}du=\int (\frac{\sec u+1}{\sec u+1}-\frac{1}{\sec u+1})du$

$=\int (1-\frac{1}{\sec u+1})du$

$=u-\int \frac{1}{\frac{{\tan }^2\frac{u}{2}+1}{1-{\tan }^2\frac{u}{2}}+1}du$

$v=\tan \frac{u}{2}$

$\Rightarrow du=\frac{2}{{\sec }^2\frac{u}{2}}$

$du=\frac{2}{v^2+1}dv$

$=u-\int \frac{(v-1)(v+1)}{v^2+1}dv$

$=u-\int \frac{v^2+1}{v^2+1}-\frac{2}{v^2+1}dv=u-\int 1-\frac{2}{v^2+1}dv$

$=u-(v-2{\tan }^{-1}(v\left)\right)$

Resubstituting $v=\tan \frac{u}{2}$

$=u-(\tan \frac{u}{2}-2{\tan }^{-1}\left(\tan \frac{u}{2}\right))$

$=\tan \frac{u}{2}$

Resubstituting $u={\sec }^1\left(x\right)$

$\tan \left(\frac{{\sec }^{-1}\left(x\right)}{2}\right)=\frac{\sqrt{x-1}}{\sqrt{x+1}}$

$\therefore \int \frac{1}{(x+1)\sqrt{x^2-1}}dx=\frac{\sqrt{x-1}}{\sqrt{x+1}}+C$.