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

Question

William Abdalla · Accepted Answer

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

#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#

Cameron Alexander · Answer

undefined To integrate $ \int x \arctan(x) $ using integration by parts:

1. Choose $ u = \arctan(x) $ and $ dv = x \, dx $.
2. Compute $ du $ and $ v $ using the derivatives and antiderivatives of $ u $ and $ dv $.
3. Apply the integration by parts formula: $ \int u \, dv = uv - \int v \, du $.
4. Substitute the values of $ u $, $ du $, $ v $, and $ dv $ into the integration by parts formula and evaluate the resulting integral.