Question

$\underset{-1}{\overset{1}{\int }}\frac{1-x^2}{1+x^2}dx=$

Answer 1

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

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

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

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

$=4{\int }_0^1\frac{1}{1+x^2}dx-2{\int }_0^{1}dx$

$=4[{\tan }^{-1}{]}_0^1-2[x{]}_0^1=4\left[{\tan }^{-1}\right[1]-{\tan }^{-1}(0\left)\right]-2[1-0]$

$=4\times \frac{\pi }{4}-2=\pi -2$.