Integration By Parts

Now you know derivatives and integrals. What is next? There are still a lot of topics you will familiarise yourself with, at least if you want to pass AP Calculus BC. But here is one of them: Integration By Parts.

Wait… What does that even mean?

Let me explain with a simple question.

How would you approach this question? There are different ways to solve this question but the one I’ll write today involves integration by parts. Integration by parts is a concept you use basically when there are two difficult functions that get multiplied in the integral.

First, let me show you the theory.

Let f(x) and g(x) be 2 continuous functions of x. Then we know that,

more conveniently,

What function to give g(x)?

Well, you should select the g(x) function giving priority over the following expressions.

The g function should first be a logarithmic function, if there isn’t any logarithmic function in the expression then you should look for an inverse trigonometric function, and so on.

Let’s go back to the question.

Instead of the g and f functions, I will call them the u and v functions.

According to the equation we found above and to the question, I will define u(x) and v(x) as the followings:

so,

Putting v(x) and u(x) back into the first equation

Now we have to calculate

in the same way. This time,

Let’s apply the same procedure.

Now we put the value back into the first equation.

Don’t forget the + C of course!

And that’s how you calculate this integral with the help of the integration by parts method. I’ll leave you with one more question and add the answer at the end of the article.

Try to solve this problem before checking the answer, please! And see you in another article. Until then, take care!

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