Integration by parts with limits is really the same as integration by parts with indefinite integrals until the very end of your calculation. Let’s see how to do integration by parts with limits with an example:

int{2}{3}{ ln(x)^2 dx}

Its always a good idea to start out writing the integration by parts formula:

int{}{}{u dv}~=~uv~-~int{}{}{v du}

For now, we can simply ignore the limits of integration. First we take

u ~=~ln(x)^2~doubleright~du=2{{ln(x)}/x}

Then we have

dv~=~dx~doubleright~v~ =~ int{}{}{ dx}~=~x

And so

int{}{}{ ln(x)^2 dx} ~=~x ln(x)^2~-~2 int{}{}{ln(x) dx}

Now we’ve got to apply integration by parts on the integral we have at the end. Let’s go through it in case you aren’t familiar with it. Take

u ~=~ln(x)~doubleright~du={1/x}

dv~=~dx~doubleright~v~ =~ int{}{}{ dx}~=~x

Hence

int{}{}{ ln(x) dx} ~=~x ln(x)~-~x~=~x(lnx~-~1)

Now we can put this together with our previous result, and we’ve found that

int{}{}{ ln(x)^2 dx} ~=~x ln(x)^2~-~2 (x(lnx~-~1))

If we were doing an indefinite integral, we would add a constant of integration to the end and we’d be done. For integration by parts with limits, all we do now is evaluate this expression at the limits of integration. The upper limit was x = 3, which gives:

3 ln(3)^2~-~2 (3(ln3~-~1))~=~3ln(3)^2~-~6 ln(3)~+~6

The lower limit was x =2, which gives
2 ln(2)^2~-~2 (2(ln2~-~1))~=~2ln(2)^2~-~4 ln(2)~+~4

So the answer is found by subtracting the lower limit from the upper limit, and you can verify that

int{2}{3}{ ln(x)^2 dx} ~=~3ln(3)^2~-~6ln(3)~-~2ln(2)^2~+~4ln(2)~+~2

So, to do integration by parts with limits, simply do the integral the way you would in the indefinite case, ignoring the limits of integration. Then when you have your answer at the end, evaluate at each limit and subtract.

Want a calculus book with step-by-step solutions? Visit calculus without limits