`int1/((x^(2)+1)^(2))`dx=…

Options

A. `tan^(-1)x-1/(2x(x^(2)+1))+c`
B. `1/2tan^(-1)x+x/(2(x^(2)+1))+c`
C. `tan^(-1)x+1/(x^(2)+1)+c`
D. `tan^(-1)x+1/(2(x^(2)+1))+c`

Answer

B

Explanation

`int1/(x^(2)+1)dx=1/(x^(2)+1)`
`int1dx-int(d/(dx)(1/(x^(2)+1))int1dx)dx`
`=x/(x^(2)+1)-int(-2x)/((x^(2)+1)^(2))xdx` (integration by parts)
`=x/(x^(2)+1)+2int(x^(2)+1-1)/((x^(2)+1)^(2))dx`
`=x/(x^(2)+1)+2intx/((x^(2)+1))dx-2int(dx)/((x^(2)+1)^(2))` .....(i)
And also we know that
`int(dx)/(x^(2)+1)=tan^(-1)(x)`
From Eqs. (i) and (ii), we get
`tan^(-1)x=x/(x^(2)+1)+2tan^(-1)x-2int(dx)/((x^(2)+1)^(2))`
`rArr2int(dx)/((x^(2)+1)^(2))=tan^(-1)x+x/(x^(2)+1)`
`rArrint(dx)/((x^(2)+1)^(2))=1/2tan^(-1)x+x/(2(x^(2)+1))=c`