Question

${\int }_{-1}^1\frac{1}{(1+x^2{)}^2}dx=?$

• A. $\frac{\pi }{4}+\frac{1}{2}$
• B. $\frac{\pi }{4}-\frac{1}{2}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{16}$

Answer 1

Answer: (A)

$\frac{\pi }{4}+\frac{1}{2}$

Concept: If f(x) is even function i.e, $f(-x)=f\left(x\right)$, then

${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$

Now
${\int }_{-1}^1\frac{1}{(1+x^2{)}^2}dx$

Let $x=tan\theta$
$dx=se{c}^2\cdot d\theta$

$={\int }_{{}^{-\pi }{/}_4}^{{}^{\pi }{/}_4}\frac{se{c}^2\theta d\theta }{(se{c}^2\theta {)}^2}$
$={\int }_{{}^{-\pi }{/}_4}^{{}^{\pi }{/}_4}{\cos }^2\theta$

$=2{\int }_0^{{}^{\pi }{/}_4}co{s}^2\theta d\theta$

$=2{\int }_0^{{}^{\pi }{/}_4}\left(\frac{1+cos2\theta }{2}\right)d\theta$

$=\frac{\pi }{4}+{\left(\frac{sin2\theta }{2}\right)}_0^{\frac{\pi }{4}}$

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