# Homework Help: [Integration] A tough substitution problem

1. Feb 1, 2012

### gunnargolf

1. The problem statement, all variables and given/known data

The following is to be evaluated using substitution or partial integration.

$$\int\frac{(2x-1)}{e^{\arctan(x)}}\,dx$$

(It's supposed to be e^(arctan(x)) but i'm new to LaTeX and can't quite figure out how I would input it correctly) (Fixed it for you.)

2. Relevant equations

No relevant equations

3. The attempt at a solution

I tried partial integration and that definitely does not help. I simply have no idea how I could substitute.

Last edited by a moderator: Feb 1, 2012
2. Feb 1, 2012

### lanedance

how about trying integration by parts first, and let u' = 2x-1

3. Feb 1, 2012

### lanedance

actually i might have to rethink that...

4. Feb 1, 2012

### susskind_leon

Imho it makes sense to substitute u=arctan(x). From there, there is still some way to go though.

5. Feb 1, 2012

### vela

Staff Emeritus
You just have to put everything in the exponent inside curly braces to group them together. Also, use a backslash before the name of common functions to get them to typeset correctly.

There's a LaTeX guide here: https://www.physicsforums.com/showthread.php?t=546968

6. Feb 1, 2012

### lanedance

or noting that $\frac{d}{dx}arctan(x)= \frac{1}{1+x^2}$ you could rearrange as follows

$$\int (2x-1)e^{-arctan(x)}= 2xe^{-arctan(x)}-1e^{-arctan(x)} = 2xe^{-arctan(x)}-\frac{1+x^2}{1+x^2}e^{-arctan(x)} = 2xe^{-arctan(x)}-(1+x^2)\frac{e^{-arctan(x)} }{1+x^2}$$

which should avoid subtitution or parts all together.. though it comes from a similar idea to parts

7. Feb 2, 2012

### susskind_leon

where did you carry out the integral?

8. Feb 2, 2012

### lanedance

haven't done the integral, as i left that part for the OP, didn't mean to have the integral sign at the start of that expression, was rearranging to simplify the integration

Last edited: Feb 2, 2012
9. Feb 2, 2012

### lanedance

So the integral becomes

$$\int \left( 2xe^{-arctan(x)}-(1+x^2)\frac{e^{-arctan(x)} }{1+x^2} \right)dx$$

10. Feb 2, 2012

### sunjin09

It can be verified that (1+x^2)/e^atan(x) is the answer, hope that helps