Question

Evaluate ${\int }_0^1\frac{dx}{(1+x^2{)}^2}$

• A. $\frac{1}{4}+\frac{\pi }{4}$
• B. $-\frac{1}{4}-\frac{\pi }{8}$
• C. $\frac{1}{4}-\frac{\pi }{4}$
• D. $\frac{1}{4}+\frac{\pi }{8}$

Answer 1

Answer: (D)

$\frac{1}{4}+\frac{\pi }{8}$

We have to find ${\int }_0^1\frac{dx}{(1+x^2{)}^2}$

Substitute $x=tanu$

Differentiating we get, $dx=se{c}^{2}udu\Rightarrow dx=\frac{1}{co{s}^{2}u}du$

Also when $x=0,u=ta{n}^{-1}x=ta{n}^{-1}\left(0\right)=0$ and when $x=1,u=ta{n}^{-1}x=ta{n}^{-1}\left(1\right)=\frac{\pi }{4}$.

$\therefore {\int }_0^1\frac{dx}{(1+x^2{)}^2}={\int }_0^{\pi /4}\frac{1}{co{s}^{2}u(1+ta{n}^{2}u{)}^2}du$

Since $ta{n}^{2}u=se{c}^{2}u-1$ we get

$={\int }_0^{\pi /4}\frac{1}{co{s}^{2}u(1+se{c}^{2}u-1{)}^2}du$

$={\int }_0^{\pi /4}\frac{1}{co{s}^{2}u(se{c}^{2}u{)}^2}du$

$={\int }_0^{\pi /4}\frac{1}{co{s}^{2}u\left(se{c}^{4}u\right)}du$

$={\int }_0^{\pi /4}\frac{co{s}^{4}u}{co{s}^{2}u}du$

$={\int }_0^{\pi /4}co{s}^{2}udu$

Since $co{s}^{2}u=\frac{1+cos\left(2u\right)}{2}$ we get

$={\int }_0^{\pi /4}\frac{1+cos\left(2u\right)}{2}du$

$=\frac{1}{2}{\int }_0^{\pi /4}(1+cos2u)du$

$=\frac{1}{2}{[u+\frac{sin2u}{2}]}_0^{\pi /4}$

$=\frac{1}{2}[\frac{\pi }{4}+\frac{sin\left(\pi /2\right)}{2}-0-0]$

$=\frac{1}{2}[\frac{\pi }{4}+\frac{1}{2}]$

$=\frac{\pi }{8}+\frac{1}{4}$

$\therefore {\int }_0^1\frac{dx}{(1+x^2{)}^2}=\frac{\pi }{8}+\frac{1}{4}$