Question

Evaluate the definite integral:

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

• A. $\frac{9}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{19}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

$I={\int }_0^1\frac{1-x^2}{(1+x^2{)}^2}dx$

Dividing by $x^2$ in both numerator and denominator

$I={\int }_0^1\frac{\frac{1}{x^2}-1}{{(x+\frac{1}{x})}^2}dx$

Let $x+\frac{1}{x}=t$.

Then, $d(x+\frac{1}{x})=dt\Rightarrow (1-\frac{1}{x^2})dx=dt$

Clearly,

$x=0\Rightarrow t=\infty$

$x=1\Rightarrow t=2$

$\therefore {\int }_{\infty }^2-\frac{1}{t^2}dt={\left[\frac{1}{t}\right]}_{\infty }^2=\frac{1}{2}$