Question

The value of $\int \frac{d(x^2+1)}{\sqrt{x^2+2}}$ is equal to

• A. $\sqrt{(x^2-2)}+C$
• B. $2\sqrt{(x^2+2)}+C$
• C. $2\sqrt{(x^2+3)}+C$
• D. $\sqrt{(x^2+2)}+C$

Answer 1

Answer: (B)

$2\sqrt{(x^2+2)}+C$

$I=\int \frac{d\left(x^2+1\right)}{\sqrt{x^2+2}}\Rightarrow \int {\left(\sqrt{\left(x^2+1\right)+1}\right)}^{-1}d\left(x^2+1\right)\Rightarrow \int {\left\{\left(x^2+1\right)+1\right\}}^{-\frac{1}{2}}d\left(x^2+1\right)\Rightarrow 2{\left(x^2+1+1\right)}^{\frac{1}{2}}+c$

$=2\sqrt{x^2+2}+C$

Hence, this is the answer.