Question

Evaluate the integral : $ \int \frac{dy}{\left(y+6\right)\left(y+5\right)^{1/2}} $ The result is

• A. $ 2 \,tan ^{-1} \left(\sqrt{\left(y+6\right)}\right) + $ constant
• B. $ 2 \,tan ^{-1}\left(y+6\right)+ $ constant
• C. $ 2\, \,tan ^{-1} \left(\sqrt{\left(y+5\right)}\right) + $ constant
• D. $ 2\, tan ^{-1}\left(y+5\right)+ $ constant

Answer 1

Answer: (C)

Let $I=\int \frac{dy}{\left(y+6\right)\sqrt{y+5}}$
put $y+5=t$ SO that $dy = dt$
$\therefore I=\int \frac{dt}{\left(t+1\right)\sqrt{t}}$
$=\int\frac{dt}{\left[\left(\sqrt{t}\right)^{2}+1\right]\sqrt{t}}$
Now put $\sqrt{t}=z$
$\Rightarrow \frac{1}{2\sqrt{t}} dt =dz$
$\therefore I=\int \frac{2dz}{z^{2}+1}$
$=2 \, tan^{-1} z+c $
$=2\, tan^{-1} \sqrt{t}+c$
$=2\,tan^{-1} \sqrt{y+5}+c$