Harder improper integral

**Glitch** · October 28th 2010, 05:15 AM

The question:
Find the $\lim_{R \to \infty} \int_{-R}^{2R} \frac{x}{1 + x^2} dx$

My attempt:
Let $u = 1 + x^2$

$du = 2x dx$

$\frac{1}{2}\int_{-R}^{2R} \frac{du}{u}$

$\frac{1}{2}_{-R}^{2R}[log(u)]$

$\frac{1}{2}[log(1+(2R)^2) - log(1 + (-R)^2)]$

$\frac{1}{2}[log(1+4R^2) - log(1+R^2)]$

$\frac{1}{2}[log(\frac{1+4R^2}{1+R^2})]$

Limit of the function inside log is 4, so:
$\frac{1}{2}log(4)$

However, the answer is log(2). Where did I go wrong? Thanks!

**TheCoffeeMachine** · October 28th 2010, 05:23 AM

Hint: [LaTeX ERROR: Convert failed]

Spoiler:

**Glitch** · October 28th 2010, 05:25 AM

Oh dear, I can't believe I got so far and tripped up on that! Thanks!

**TheCoffeeMachine** · October 28th 2010, 05:28 AM

Originally Posted by Glitch

Oh dear, I can't believe I got so far and tripped up on that! Thanks!

Haha! It happens to us all.

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