Question

Integration of : `int(-1/2x+1/2)/(x^2 +1)dx` ?

`int(-1/2x+1/2)/(x^2 +1)dx`

Answer 1

`int (-1/2x+1/2)/(x^2+1) dx = 1/2 arctanx -1/4 ln(1+x^2)+C`

Explanation:

Simplify the integrand:

`int (-1/2x+1/2)/(x^2+1) dx = 1/2 int (1-x)/(x^2+1) dx`

using the linearity of the integral:

`int (-1/2x+1/2)/(x^2+1) dx = 1/2 int dx/(1+x^2) -1/2 int (xdx)/(1+x^2)`

solve the two integrals:

`int dx/(1+x^2) = arctanx +c_1`

`int (xdx)/(1+x^2) = 1/2 int (d(1+x^2))/(1+x^2) = 1/2 ln (1+x^2) +c_2`

and substituting the partial solutions:

`int (-1/2x+1/2)/(x^2+1) dx = 1/2 arctanx -1/4 ln(1+x^2)+C`