Question

Integral of $\frac{1}{\sqrt{x^2+4}}$ with respect to $(x^2+3)$ is equal to

• A. $\sqrt{x^2+4}+C$
• B. $\frac{1}{\sqrt{x^2+4}}+C$
• C. $2\sqrt{x^2+4}+C$
• D. None of these

Answer 1

The correct option is C $2\sqrt{x^2+4}+C$

$\int \frac{d(x^2+3)}{\sqrt{x^2+4}}=\int \frac{d(x^2+3)}{\sqrt{(x^2+3)+1}}=\int \frac{dt}{\sqrt{t+1}}$
[Put $x^2+3=t\Rightarrow d(x^2+3)=dt$]
$=2\sqrt{t+1}+C=2\sqrt{x^2+4}+C$