1. ## Integration of x/ax^2+c

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.

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

3. 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.

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

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

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

6. 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?

7. 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.

8. 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?

9. 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.

10. 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 $\int\frac{x dx}{ax^2+ 8}= \int\frac{\frac{1}{2a}du}{u}= \frac{1}{2a}\int \frac{du}{u}$