Question

The solution of $\int \frac{dx}{\sqrt{x^2+c}}$ using Euler's substitution is?

• A. $log\mid x-\sqrt{x^2+c}\mid$
• B. $log\mid x+\sqrt{x^2+c}\mid$
• C. $log\mid x+\sqrt{x^2-c}\mid$
• D. $log\mid x-\sqrt{x^2-c}\mid$

Answer 1

Answer: (B)

$log\mid x+\sqrt{x^2+c}\mid$

We have to find the solution of $\int \frac{dx}{\sqrt{x^2+c}}$ using Euler substitution.
Consider $\int \frac{dx}{\sqrt{x^2+c}}$
We can use Euler first substitution: $\sqrt{x^2+c}=-x+t$
$\Rightarrow x^2+c=(-x+t{)}^2$
$\Rightarrow x^2+c=x^2-2xt+t^2$
$\Rightarrow x=\frac{t^2-c}{2t}$
Differentiating both sides we get
$dx=\frac{t^2+c}{2t^2}dt$
Also $\sqrt{x^2+c}=-\frac{t^2-c}{2t}+t=\frac{t^2+c}{2t}$
$\therefore \int \frac{dx}{\sqrt{x^2+c}}=\int \frac{\frac{t^2+c}{2t^2}dt}{\frac{t^2+c}{2t}}$
$=\int \frac{dt}{t}$
$=log|t|$
$=log|x+\sqrt{x^2+c}|$
Hence $\int \frac{dx}{\sqrt{x^2+c}}=log|x+\sqrt{x^2+c}|$