Question

How do you integrate `int xarctanx` by integration by parts method?

Answer 1

`((x^2-1)arctan(x)-x)/2+C`

Explanation:

`I=intxarctan(x)dx`

Integration by parts takes the form `intudv=uv-intvdu`. So, for `intudv=intxarctan(x)dx`, we should let:

`{(u=arctan(x),=>,du=dx/(1+x^2)),(dv=xdx,=>,v=x^2/2):}`

Thus:

`I=1/2x^2arctan(x)-1/2intx^2/(1+x^2)dx`

Rewrite the numerator or perform polynomial long division of the integrand. Both will result in equivalent simplifications:

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

`I=1/2x^2arctan(x)-1/2int(1+x^2)/(1+x^2)dx-1/2int1/(1+x^2)dx`

`I=1/2x^2arctan(x)-1/2intdx-1/2int1/(1+x^2)dx`

Both of these are simply integrated:

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

Or:

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