Question

Evaluate $\int \frac{dx}{1+\sqrt{x^2+2x+2}}$.

• A. $I=\ln (x+1-\sqrt{x^2+2x+2})+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$
• B. $I=\ln (x-2-\sqrt{x^2-2x-4})+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$
• C. $I=\ln (x+1+\sqrt{x^2+2x+2})+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$
• D. None of these

Answer 1

The correct option is C $I=\ln (x+1+\sqrt{x^2+2x+2})+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$

$\sqrt{x^2+2x+2}=t-x$
Squaring both the sides, we get
$x^2+2x+2=t^2+x^2-2tx$
$2x+2tx=t^2-2$
$x=\frac{t^2-2}{2(1+t)}$
$\Rightarrow dx=\frac{t^2+2t+2}{2{(1+t)}^2}dt$
$\therefore 1+\sqrt{x^2+2x+2}=1+t-\frac{t^2-2}{2(1+t)}=\frac{t^2+4t+4}{2(1+t)}$
$⟹I=\int \frac{2(1+t)(t^2+2t+2)}{(t^2+4t+4)2{(1+t)}^2}dt$
$I=\int \frac{(t^2+2t+2)}{(1+t){(t+2)}^2}dt$
Now resolving into partial fractions,
$\frac{t^2+2t+2}{(1+t){(t+2)}^2}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{{(t+2)}^2}$
$\frac{t^2+2t+2}{(1+t){(t+2)}^2}=\frac{A{(t+2)}^2+B(t+2)(t+1)+C(t+1)}{t+1}$
$\Rightarrow t^2+2t+2=A{(t+2)}^2+B(t+2)(t+1)+C(t+1)$
Put $t=-1\Rightarrow A=1$
Put $t=-2\Rightarrow C=-2$
Put $t=0\Rightarrow B=0$
So, $I=\int \frac{dt}{t+1}-2\int \frac{dt}{{(t+2)}^2}$
$=\ln |t+1|+\frac{2}{(t+2)}+C$
$I=\ln (x+1+\sqrt{x^2+2x+2})+\frac{2}{(x+2)+\sqrt{x^2+2x+2}}+C$