I need to find the Integral of Sqrt(x^2+2x)dx. It is a Integration by Partial Fractions exercise.

I don't even know where to start. Any help on where to start would be greatly appreciated.

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- Feb 5th 2009, 05:01 PMdjo201Urgent Integration Help
I need to find the Integral of Sqrt(x^2+2x)dx. It is a Integration by Partial Fractions exercise.

I don't even know where to start. Any help on where to start would be greatly appreciated. - Feb 5th 2009, 05:06 PMKrizalid
I don't see the way of applying partial fractions here. Do we have a typo?

- Feb 5th 2009, 05:08 PMdjo201
- Feb 5th 2009, 05:11 PMmr fantastic
- Feb 5th 2009, 05:12 PMKrizalid
Then integrate by parts or use a trig. substitution.

- Feb 5th 2009, 05:16 PMdjo201
Not a typo there neither. It's the integral of the square root of (x^2 + 2x) dx

Krizalid: If it isn't necessary I don't have to use trigonometric expressions. If it's the only way, please explain a bit further.

EDIT: If this helps, the solution given by the book is:

(1/2)[(x+1)Sqrt(x^2+2x)-ln|x+1+Sqrt(x^2+2x)|]+c - Feb 5th 2009, 05:32 PMdjo201
Sorry for Double Post.

I got confused. It's not from the Partial Fractions Section. It's a Trig. Substitution exercise. - Feb 5th 2009, 06:09 PMKrizalid
I know something was wrong.

Well then, to perform a trig. substitution, we have $\displaystyle x^2+2x=x^2+2x+1-1=(x+1)^2-1$ so you need to put $\displaystyle x+1=\sec\varphi.$

(I closed your duplicate thread on calculus section, there's no reason to make an identical one.) - Feb 5th 2009, 06:12 PMdjo201
- Feb 5th 2009, 06:15 PMKrizalid
Yes, the first one is doable by parts and the second one requires some clever solution:

$\displaystyle \sec z=\frac{{{\sec }^{2}}z+\sec z\tan z}{\sec z+\tan z}.$

Note that by differentiating the denominator you get the numerator. - Feb 5th 2009, 06:23 PMdjo201
- Feb 5th 2009, 06:27 PMKrizalid
$\displaystyle \int\sec^3z\,dz=\int\sec z\sec^2z\,dz=\int\sec z(\tan z)'\,dz.$

Strongly recommended lecture: make your partial integration faster. (See my signature.)

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Or if you want to use the classic way, put $\displaystyle u=\sec z$ and $\displaystyle dv=\sec^2z\,dz.$ - Feb 5th 2009, 06:45 PMdjo201
- Feb 5th 2009, 06:48 PMKrizalid
No, I'm not gonna kill ya 'cause you're doing a good work. Now observe that,

$\displaystyle \sec z\tan^2z=\sec z(\sec^2z-1)=\sec^3z-\sec z.$

Is this familiar few posts ago? - Feb 5th 2009, 06:53 PMdjo201