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

1 Answer
mason m
Nov 6, 2016

((x^2-1)arctan(x)-x)/2+C(x2−1)arctan(x)−x2+C

Explanation:

I=intxarctan(x)dxI=∫xarctan(x)dx

Integration by parts takes the form intudv=uv-intvdu∫udv=uv−∫vdu. So, for intudv=intxarctan(x)dx∫udv=∫xarctan(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

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