Question

How do you find the integral `ln(x^2+4)`?

Answer 1

`I = xln(x^2+4)- 2x + 4arctan(x/2)+C`

Explanation:

`I=int ln(x^2+4)dx`

`u=ln(x^2+4) => du=1/(x^2+4)*2x*dx= (2x)/(x^2+4)dx`

`dv=dx => v=x`

`I=xln(x^2+4)- int x(2x)/(x^2+4)dx`

`I = xln(x^2+4)- 2int x^2/(x^2+4)dx`

`I = xln(x^2+4)- 2int (x^2+4-4)/(x^2+4)dx`

`I = xln(x^2+4)- 2int (1-4/(x^2+4))dx`

`I = xln(x^2+4)- 2(x-4*1/2arctan(x/2))+C`

`I = xln(x^2+4)- 2x + 4arctan(x/2)+C`