Integration Problem Has Me Stumped

**Sucker Punch** · July 4th 2010, 12:35 AM

One question on my assignment has me stumped. I've tried a number of different methods, but every time, I either get stuck somewhere, or the equation quickly gets ridiculously out of hand. I'm not sure where to start here.

$\int \frac {x^2}{\sqrt[]{2x^2+4x+25}}dx$

I've used google and some math help sites to try to point myself in the right direction, but it hasn't been much help. I'm getting answers like this;

Which both seem very wrong, and don't help at all.

Can someone give me some advice?

**simplependulum** · July 4th 2010, 12:46 AM

Hi , so what is your answer , let's compare it with the answers you have got in the sites .

Let me show that the two results are the same :

We can see that both of them have this thing :

$\frac{1}{8} [ 2(x-3) \sqrt{2x^2 + 4x + 25 }]$

so let's compare $\log\left( \frac{ \sqrt{2}x +\sqrt{2} }{\sqrt{23} } + \sqrt{ \frac{( \sqrt{2}x +\sqrt{2})^2}{23} + 1 }\right)$ and $\sinh^{-1}\left( \sqrt{\frac{2}{23}} (x+1) \right)$

In fact , we have $\sinh^{-1}(u) = \ln( u + \sqrt{u^2 + 1 } )$

by letting $u = \sqrt{\frac{2}{23}} (x+1) = \frac{ \sqrt{2}x +\sqrt{2} }{\sqrt{23}}$

We have $\log\left( \frac{ \sqrt{2}x +\sqrt{2} }{\sqrt{23} } + \sqrt{ \frac{( \sqrt{2}x +\sqrt{2})^2}{23} + 1} \right) = \sinh^{-1}\left( \sqrt{\frac{2}{23}} (x+1) \right)$ so actually the two answers you've got are the same .

**Also sprach Zarathustra** · July 4th 2010, 12:53 AM

Press the "Show steps": integrate x^2/(sqrt(2x^2+4x+25&# 41;) - Wolfram|Alpha.

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