Question

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

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

Answer 1

The correct option is B $\frac{2}{\sqrt{3}}\left(ta{n}^{-1}\right(\frac{\sqrt{x-2}}{\sqrt{3}}\left)\right)+C$

The given form is one of the forms of irrational algebraic functions which is $\int \frac{1}{(ax+b)\sqrt{cx+d}}dx.$
Now to evaluate such kind of integrals we need to substitute $cx+d=t^2$
So, here we’ll substitute $x-2=t^2$
& dx = 2t.dt
So the integral becomes
$\int \frac{2t.}{(t^2+3)t.}dt$
$=\int \frac{2}{(t^2+3)}dt$ (use the standard formulae)
$=\frac{2}{\sqrt{3}}\left(ta{n}^{-1}\right(\frac{t}{\sqrt{3}}\left)\right)+C$
Or $\frac{2}{\sqrt{3}}\left(ta{n}^{-1}\right(\frac{\sqrt{x-2}}{\sqrt{3}}\left)\right)+C$ (Substituting t = $\sqrt{x-2}$)