Question

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

• A. $\log |\frac{\sqrt{x+2}+1}{\sqrt{x+2}-1}|+c$
• B. $\log |\frac{\sqrt{x+2}-1}{\sqrt{x+2}+1}|+c$
• C. $\frac{1}{5}\log |\frac{\sqrt{x+2}-1}{\sqrt{x+2}+1}|+c$
• D. $\frac{1}{5}\log |\frac{\sqrt{x+2}+1}{\sqrt{x+2}-1}|+c$

Answer 1

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

Let $I=\int \frac{1}{(x+1)\sqrt{x+2}}dx$

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

$\sqrt{x+2}=t=\frac{1}{2\sqrt{x+2}}dx=dt$ and $x+2=t^2$

$\therefore I=\int \frac{2dt}{(t^2-1)}=2\int \frac{dt}{t^2-1}$

$=\frac{2}{2}log\mid \frac{t-1}{t+1}\mid +c$

Now, after putting the value of $t$, we get

$\int \frac{1}{(x+1)\sqrt{x+2}}dx=log\mid \frac{\sqrt{x+2}-1}{\sqrt{x+2}+1}\mid +c$