Question

The value of the integral ${\int }_0^1\frac{1}{(x^2+1{)}^{3/2}}dx$ is

• A. $\frac{3}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $1$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{2}}$

Let $I=\int \frac{1}{{(x^2+1)}^{\frac{3}{2}}}dx$
Substituting $x=tant\Rightarrow dx=se{c}^{2}tdt$, we get
$I=\int cot\left(t\right)dt=sin\left(t\right)=sin\left(ta{n}^{-1}x\right)=\frac{x}{\sqrt{x^2+1}}$
Therefore,
${\int }_0^1\frac{1}{{(x^2+1)}^{\frac{3}{2}}}dx={\left[\frac{x}{\sqrt{x^2+1}}\right]}_0^1=\frac{1}{\sqrt{2}}$