# Evaluating the integral as a power series Tips and tricks [closed]

The purpose of this post is to show that there is more than one way to solve a problem. So I proposed one way of solving a problem shown below. I was wondering if anyone has a different way to solve this problem? Is there a more advanced way to go about solving this integral?

The goal: What I hoped to accomplish by asking this question was to engage newer users with a question they might have seen before on an exam or homework assignment.

Integration is hard but by using a power series you can evaluate more complex integrals. For example: integrating x-arctan(x)/(x^2) can be quite tricky. However, if you can put it in the form 1/1-r you can approximate the integral using a power series. From personal experience, this is a very important concept in computer science and algorithm design.

Here is a problem I solved using a power series: If you are new to calculus then the first problem is always how do I start this? First: (Think outside the box) what can be represented as 1/1-r

• In this case, it is arctan(x) By taking the derivative of arctan(x) you get 1/1+x^2

If you need a quick refresher on doing this I found this to be a great reference: https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-15-implicit-differentiation-and-inverse-functions/MIT18_01SCF10_Ses15b.pdf

Things to take note of in this question: - It is not completely in the form 1/1-r but it is close - There are no transformations to the radius of convergence yet.

Now that we have a summation we have to take the integral so that we can plug the summation back into the original equation. When integrating note everything that does not have an x is a constant.(When integrating in terms of d/dx) Great now that we have integrated we plug the summation in for arctan(x). it may seem messy at first but it will simplify as time goes on.

Looking at the summation you can see that when the first term is plugged in then we should get -x. we can also steal a negative sign and place it into the main equation, making it easy to work with.

Being the first term cancels out with x we are left with x^-2 which can be placed into the summation. Lastly, We integrate the summation.

Here is the work to go along with the thought process above

Here is the work to go along with the thought process above page 2

## closed as unclear what you're asking by Xander Henderson, Saad, user284331, Math1000, Mohammad Riazi-KermaniMar 30 '18 at 1:33

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