Question

${\int }_0^1\frac{x^2}{1+x^2}$ dx is equal to

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

Answer 1

The correct option is D $1-\frac{\pi }{4}$

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

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

${\int }_0^{1}1.dx-{\int }_0^1\frac{1}{1+x^2}dx\Rightarrow {[x-ta{n}^{-1}x]}_0^1$

$\Rightarrow (1-ta{n}^{-1}1)-(0-0)$

$=1-\frac{\pi }{4}$