Urgent Integration Help

**djo201** · Feb 5th 2009, 05:01 PM

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.

**Krizalid** · Feb 5th 2009, 05:06 PM

I don't see the way of applying partial fractions here. Do we have a typo?

**djo201** · Feb 5th 2009, 05:08 PM

Originally Posted by Krizalid

I don't see the way of applying partial fractions here. Do we have a typo?

Not a typo. That's how it's on the book. Is there any other way to solve it?

**mr fantastic** · Feb 5th 2009, 05:11 PM

Originally Posted by djo201

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.

Maybe a typo in the question ....? 1/(x^2 + 2x) .....?

**Krizalid** · Feb 5th 2009, 05:12 PM

Then integrate by parts or use a trig. substitution.

**djo201** · Feb 5th 2009, 05:16 PM

Originally Posted by mr fantastic

Maybe a typo in the question ....? 1/(x^2 + 2x) .....?

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

**djo201** · Feb 5th 2009, 05:32 PM

Sorry for Double Post.

I got confused. It's not from the Partial Fractions Section. It's a Trig. Substitution exercise.

**Krizalid** · Feb 5th 2009, 06:09 PM

I know something was wrong.

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

(I closed your duplicate thread on calculus section, there's no reason to make an identical one.)

**djo201** · Feb 5th 2009, 06:12 PM

Originally Posted by Krizalid

I know something was wrong.

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

(I closed your duplicate thread on calculus section, there's no reason to make an identical one.)

Thanks, I made the trig. subs. and I'm at the point where I have
Integral of sec^3(z)-sec(z). I'm not allowed to use the Table of Integrals. Any other way of moving on from that?

**Krizalid** · Feb 5th 2009, 06:15 PM

Yes, the first one is doable by parts and the second one requires some clever solution:

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

Note that by differentiating the denominator you get the numerator.

**djo201** · Feb 5th 2009, 06:23 PM

Originally Posted by Krizalid

Yes, the first one is doable by parts and the second one requires some clever solution:

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

Note that by differentiating the denominator you get the numerator.

Sorry for asking so many questions. But this problem is eating me. I'm trying to integrate by parts but can't find how. Do I use u=sec(z) or u=tan(z)?

Again, sorry for asking and asking. And thanks for all the help.

**Krizalid** · Feb 5th 2009, 06:27 PM

$\int\sec^3z\,dz=\int\sec z\sec^2z\,dz=\int\sec z(\tan z)'\,dz.$

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

---

Or if you want to use the classic way, put $u=\sec z$ and $dv=\sec^2z\,dz.$

**djo201** · Feb 5th 2009, 06:45 PM

Originally Posted by Krizalid

$\int\sec^3z\,dz=\int\sec z\sec^2z\,dz=\int\sec z(\tan z)'\,dz.$

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

---

Or if you want to use the classic way, put $u=\sec z$ and $dv=\sec^2z\,dz.$

I'm sure your going to kill me if you read this. I'm trying with partial integration but I'm stuck (AGAIN).

I got to sec(z)tan(z)-Integral of sec(z)tan^2(z)

Am I doing something wrong?

**Krizalid** · Feb 5th 2009, 06:48 PM

No, I'm not gonna kill ya 'cause you're doing a good work. Now observe that,

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

Is this familiar few posts ago?

**djo201** · Feb 5th 2009, 06:53 PM

Originally Posted by Krizalid

No, I'm not gonna kill ya 'cause you're doing a good work. Now observe that,

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

Is this familiar few posts ago?

Yes, it's the same as before I started to integrate by parts. If I keep trying to integrate that it's going to keep repeating and repeating. What should I do?

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