Question

Evaluate $\int \frac{1}{(x-3)\sqrt{x+1}}dx.$

• A. $\frac{1}{2}\log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|$
• B. $\frac{1}{2}\log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+C$
• C. $\frac{1}{2}\log \left|\frac{\sqrt{x-1}-2}{\sqrt{x-1}+2}\right|+C$
• D. $\frac{1}{2}\log \left|\frac{\sqrt{x+1}-2}{\sqrt{x-1}+2}\right|+C$

Answer 1

The correct option is B $\frac{1}{2}\log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+C$

Let $I=\int \frac{1}{(x-3)\sqrt{x+1}}dx$
Let $x+1=t^2$ or $dx=2tdt$
$\therefore I=\int \frac{1}{(t^2-1-3)}\frac{2t}{\sqrt{t^2}}dt$
$=2\int \frac{dt}{t^2-2^2}=2\times \frac{1}{2\left(2\right)}\log \left|\frac{t-2}{t+2}\right|+C$
$=\frac{1}{2}\log \left|\frac{\sqrt{x+1}-2}{\sqrt{x+1}+2}\right|+C$