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I am having trouble with this integral: $$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx.$$ One obvious thing would be to complete the square: $x^2+2x+25=(x+1)^2+24$. But then, I don't know which substitution to use. Can anyone help? Thank you.

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Try $u=x+1$ followed by a trig substitution. –  Eric Gregor May 6 '12 at 19:09

3 Answers 3

up vote 6 down vote accepted

I believe the substitution you're looking for is

$x+1=2\sqrt6\tan\theta,x=2\sqrt6\tan\theta-1$.

This will reduce the denominator to

$\sqrt{(2\sqrt6\tan\theta)^2+24}=\sqrt{24\tan^2\theta+24}=\sqrt{24\sec^2\theta}=2\sqrt6\sec\theta$

I assume you'll be able to take it from there?

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Thanks! This was easier than I thought! –  gotan123 May 7 '12 at 0:01

You will have, by substitution of $y=\frac{x+1}{\sqrt{24}}$,

$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy$$

that becomes

$$\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy=\sqrt{24}\int_\frac{1}{\sqrt{24}}^1\frac{y}{\sqrt{y^2+1}}dy-\int_\frac{1}{\sqrt{24}}^1\frac{1}{\sqrt{y^2+1}}dy.$$

This gives finally

$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\sqrt{48} - \sqrt{26} + {\rm arcsinh}\left(\frac{1}{\sqrt{24}}\right) - {\rm arcsinh}(1).$$

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You seem to have skipped a few steps, but this looks a bit off. Your denominator looks like $\sqrt{(5y)^2+24}$ and $x=5y-1$. –  Mike May 6 '12 at 19:44
    
@Mike: As pointed out by OP in the question $x^2+2x+25=(x+1)^2+24$ and so by using my substitution you get the result. Please, be careful before downvote. Thanks. –  Jon May 6 '12 at 20:25
    
@Jon You have only partially corrected your mistakes. It should be $\sqrt{24}y-1$ and not $\sqrt{24}(y-1)$ –  Kirthi Raman May 6 '12 at 20:29
    
It wasn't me who downvoted you. Looks like I got downvoted as well. It looks like I made a mistake of my own which has now been corrected. And I still don't know why you have that $y-1$ in parentheses. –  Mike May 6 '12 at 20:31
    
@KVRaman: Thanks a lot. You are right of course. –  Jon May 6 '12 at 21:20

We can do this in still another way (what I might refer to as the 'naive' way - set everything under the radical to a variable and go). You started with

$$\int_0^4\frac{xdx}{\sqrt{x^2+2x+25}} = \int_0^4 \frac{xdx}{\sqrt{(x+1)^2 + 24}}$$

Let $u = (x+1)^2 + 24$, so that $du = (2x+2)dx$

Then $\displaystyle \int_0^4 \frac{xdx}{\sqrt{(x+1)^2 + 24}} = \frac{1}{2}\int_0^4 \frac{2x + 2}{\sqrt{(x+1)^2 + 24}}dx - \int_0^4 \frac{dx}{\sqrt{(x+1)^2+ 24}}$

The second integral is a standard arcsinh integral, and with the change of variables above the first integral becomes $\displaystyle \frac{1}{2}\int \frac{du}{\sqrt u}$

I always like it when an integral can be computed in many ways.

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