Integration of x/ax^2+c

**Googl** · May 31st 2011, 08:40 AM

Hi,

I am revising on Solids of revolution and I am stuck on integrating a particular function. I have been given examples examples of similar variations but not this one, so I have been stuck all.

How would you integrate functions in this form:

I have been shown when the top x is a number such as 2 or 1.

Thanks. Sorry I can't type maths yet.

**FernandoRevilla** · May 31st 2011, 08:44 AM

Hint: $(\ln |u|)'=u'/u$

**Googl** · May 31st 2011, 08:52 AM

Originally Posted by FernandoRevilla

Hint: $(\ln |u|)'=u'/u$

Was that: ln (u) = 1/u ?

I understand that.

I also know that ∫1/ax+b dx = 1/a ln|ax +b|+c

The only problem is when where there is a number there is x as in my first example.

**Plato** · May 31st 2011, 09:02 AM

What is the derivative of $\frac{\ln(ax^2+8)}{2a}~?$

**FernandoRevilla** · May 31st 2011, 09:04 AM

$\int\dfrac{x\;dx}{ax^2+8}=\dfrac{1}{2a}\int\dfrac{ 2ax\;dx}{ax^2+8}=\dfrac{1}{2a}\ln |ax^2+8|+C$

Edited: Sorry, I didn't see Plato's post.

**Googl** · May 31st 2011, 09:23 AM

Originally Posted by FernandoRevilla

$\int\dfrac{x\;dx}{ax^2+8}=\dfrac{1}{2a}\int\dfrac{ 2ax\;dx}{ax^2+8}=\dfrac{1}{2a}\ln |ax^2+8|+C$

Edited: Sorry, I didn't see Plato's post.

Thanks.

How did you come up with 1/2a? isn't it supposed to be just 1/a. What happened to the x on top?

**Deveno** · May 31st 2011, 09:31 AM

what is the derivative of $ax^2+8$ ?

it's $2ax$ , is it not? but we only have x on top, so we put 2a inside the integral, and 1/(2a) outside.

**Googl** · May 31st 2011, 09:49 AM

Originally Posted by Deveno

what is the derivative of $ax^2+8$ ?

it's $2ax$ , is it not? but we only have x on top, so we put 2a inside the integral, and 1/(2a) outside.

So you're saying that when ever we have x on top we must multiply it with the derivative of the denominator at the bottom? Then multiple the current inside integral with the derivative of the denominator?

**Googl** · June 9th 2011, 03:48 AM

Originally Posted by FernandoRevilla

$\int\dfrac{x\;dx}{ax^2+8}=\dfrac{1}{2a}\int\dfrac{ 2ax\;dx}{ax^2+8}=\dfrac{1}{2a}\ln |ax^2+8|+C$

Edited: Sorry, I didn't see Plato's post.

Sorry, I asked a stupid question previously. I now see how that is formed after revising Integration by inspection. The two is actually a factor of the differentiation of the lower part of the function. We divide that by 1 or the all expression to great rid of it when it is being differentiated.

**HallsofIvy** · June 9th 2011, 04:22 AM

Perhaps you could see it better as substitution: let $u= ax^2+ 8$ . Then du= 2a x dx. Now there are two ways to think:
Rewrite the integral as $\frac{1}{2a}\int \frac{2ax dx}{ax^2+ 8}= \frac{1}{2a}\int \frac{du}{u}$

or rewrite du= 2ax dx as $\frac{1}{2a}du= dx$ so that $\int\frac{x dx}{ax^2+ 8}= \int\frac{\frac{1}{2a}du}{u}= \frac{1}{2a}\int \frac{du}{u}$

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