Young woman in a thoughtful pose standing in front of a blackboard that is covered in equations

Integration techniques: Integration by Parts, Example 2

This example probably should have been example one because it is more straightforward and therefore easier. For this example, we will integrate

∫x·ex dx

using the formula for integration by parts:

u dv = uv – ∫v du

(If you need an overview of the steps, please visit my page that explains how to do integration by parts.)

Step 1: Choose u and dv

Looking at our equation, x looks like a good option for u because it is easy to differentiate.

dv is then whatever is left that is not u; in this case, it is ex.

x·ex dx

Step 2: Calculate du and v

If u = x, then taking the derivative of both sides of the equal sign gives us
du = 1 dx = dx

dv is ex dx
v = ∫ex dx = ex + C

(Remember to add the constant! Personally, I would wait until the end to add it, but I have been told that at least one university marks off if +C isn’t carried down through all of the sub-answers to the final answer.)

To recap:

Choose: Choose:
u = x dv = ex dx
Calculate du by taking the derivative of u: Calculate v by integrating dv:
du = dx  v = ex + C

Tip: Use a table like this to keep track of your work when solving problems like this.

Step 3: Apply the Integration by Parts Formula

You now have the four pieces you needed to find in order to use the Integration by Parts formula. Plug u, du, v, and dv into the formula, and what you get should (hopefully!) be much easier to solve than the original integral. (If it isn’t easier, check to see if a different technique like Integration by Substitution might be more appropriate.)

Now we can plug into our formula for Integration by Parts:

              ∫u dv = uv − ∫v du

(1)     ∫x·ex dx = x(ex) − ∫(ex) (dx) + C

We’re getting closer to a solution, but we aren’t there yet. Now we need to solve the new integral.

Step 4: Solve the New Integral

We now have a new integral on the right side of the equationthat we need to solve. This one is easy, because the integral of ex is just itself plus a constant.

(2)       ∫ex dx = ex + C

Step 5: Put the pieces together

Remember: After writing down your final answer, always tack on a ” + C ” for the unknown constant to the end of your answer.

Here’s a summary of our partial results from the steps above:

(1)  ∫x·ex dx = x(ex) − ∫(ex) (dx) + C

(2)  ∫(ex )(dx) = ex + C

Substituting our new equation (2) into (1) will give us our correct final answer:

∫x·ex dx = x·ex− ex + C

This is our final answer:

∫x·ex dx = x·ex− ex + C

As you can see from this, Integration by Parts isn’t always as tedious as the first example. Nevertheless, be sure to leave plenty of time to work on these problems, though, because you never know whether you will get one like this or one like the first example.

I will be adding another example of Integration by Parts next, followed by a series on Integration by Substitution. If you have questions about this or think of anything you would like to see covered, please leave a comment below! ↓