Integration by Parts of Inverse Tangent

[James98765]

Homework Statement

I must evaluate the indefinite integral:
\int x \arctan{x} dx

Homework Equations

I am using the following format to perform the integration:
\int u dv = uv - \int v du

The Attempt at a Solution

I have tried working the problem substituting x in for u and arctan in for dv and here is my work:

Let \ u=x; \ du = dx
Let \ v = \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}; \ dv = \arctan{x}
\int x \arctan{x} dx = x (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) - \int (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) dx
2 \int \arctan{x} dx = x (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) + \int \frac{1}{2} \ln{(1 + x^2)} dx

I then apply another integration by parts technique:

Let \ u = \frac{1}{2} \ln{(1 + x^2)}; \ Let \ du = \frac{x}{1 + x^2} dx
Let \ v = x; \ Let \ dv = dx
\int \frac{1}{2} \ln{(1 + x^2)} dx = \frac{1}{2} x \ln{(1 + x^2)} - \int \frac{x}{1 + x^2} dx
\int \frac{1}{2} \ln{(1 + x^2)} dx = \frac{1}{2} x \ln{(1 + x^2)} - \frac{1}{2} \ln{(1 + x^2)}

I then substitute the second integration into the first and obtain:

\int x \arctan{x} dx = x (x \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}) - \frac{1}{2} x \ln{(1 + x^2)} - \frac{1}{2} \ln{(1 + x^2)}
2 \int x \arctan{x} dx = x^2 \arctan{x} - \frac{1}{2} \ln{(1 + x^2)}
\int x \arctan{x} dx = \frac{1}{2} x^2 \arctan{x} - \frac{1}{4} \ln{(1 + x^2)} + C

The answer in the back of the book is:

\int x \arctan{x} dx = \frac{1}{2} (x^2 + 1) \arctan{x} - \frac{1}{2} x + C

I just can't seem to find out what I am doing wrong with my integration. I graphed both the book's answer and my answer on a graphing calculator and found my answer to be wrong by a little but not by much. Could somebody show me where my technique is flawed? I don't see how the book's answer doesn't contain any natural logarthmic functions. Thank you!
-James

 

[Дьявол]

Ok, obviously you messed up the things.

First let u=arctan(x), and dv=x

What are du and v?

What do you think you should do next?

Regards.

 

[James98765]

Thanks for your help. Just after you responded I found my mistake. I have tried using the substitution you recommended before but I kept getting the wrong answer because of a small math error. Now I have found it. Thanks!
-James

 

[Cyosis]

Ok, obviously you messed up the things.

First let u=arctan(x), and v=x

What are du and dv?

What do you think you should do next?

Regards.

The mistake is not so obvious in fact although cumbersome his first partial integration was correct. Also using your suggestion you would get \int \arctan x dx yet the integral he needs to solve is \int x\arctan x dx. The best way to do this would be u=arctan(x) and v=1/2 x^2.That said the OPs method works as well it is just more difficult. This is where you made the error:
\int \frac{1}{2} \ln{(1 + x^2)} dx = \frac{1}{2} x \ln{(1 + x^2)} - \int \frac{x}{1 + x^2} dx

It should be:
<br /> \frac{1}{2} x \ln{(1 + x^2)} - \int \frac{x^2}{1 + x^2} dx

 

[James98765]

Thank you. That helps a lot. I have been looking at that for a long time and just couldn't find the mistake.

 

[Дьявол]

The mistake is not so obvious in fact although cumbersome his first partial integration was correct. Also using your suggestion you would get \int \arctan x dx yet the integral he needs to solve is \int x\arctan x dx. The best way to do this would be u=arctan(x) and v=1/2 x^2.

Yep, sorry I made typo. I meant u=arctan(x) and dv=x, so that v=x^2/2

Regards.

Thank you. That helps a lot. I have been looking at that for a long time and just couldn't find the mistake.

You're welcome!