integrate (4x^2)/(x^3+2)
where do I even start? I know part of it is going to be ln (x^3+2)...

Question

integrate (4x^2)/(x^3+2)
where do I even start?  I know part of it is going to be ln (x^3+2)...

anonymous · Answer

You get really lucky in  this one, because you can do the $u$ sub on the denominator.

anonymous · Answer

how would i know to do that?

anonymous · Answer

$$
d(x^3+2)=3x^2dx
$$And so $$
\int \frac{4x^2}{x^3+2}dx = \int \frac{4}{3}\frac{1}{x^3+2}(3x^2dx)= \frac{4}{3}\int   \frac{1}{x^3+2}d(x^3+2)
$$

anonymous · Answer

I end up with 4/3 ln (x^3+2) but how do I know to u sub or try part or trig sub or something else?

anonymous · Answer

$$
\frac{4}{3}\int   \frac{1}{x^3+2}d(x^3+2)
$$Is equivalent to saying: $$
\frac{4}{3}\int   \frac{1}{u}du
$$where $u = x^3+2$

anonymous · Answer

There are many different things to rule out, so it is very complicated.

anonymous · Answer

Do $u$ sub when you think you have a derivative of another function.

anonymous · Answer

=-).....hehe yup
I think the complicated is what throws me off

anonymous · Answer

Look at the derivative of the denominator, or the derivative of the expression in the square root.

anonymous · Answer

If you see it, then you know to use $u$ sub.

anonymous · Answer

If you have $$
\int \frac{c}{a\pm bx^2}dx
$$Or $$
\int c\sqrt{a\pm bx^2}dx
$$Or even $$
\int \frac{c}{\sqrt{a\pm bx^2}}dx
$$You gotta use trig sub because the derivative of $a\pm bx^2$ is $\pm bx$ and you don't have an $x$ to factor out.

anonymous · Answer

I guess what I really need is a little algorithm to apply when I'm first given the problems - if you pick the wrong integration method it is almost hopeless

anonymous · Answer

That algorithm would NOT be little.

anonymous · Answer

Just start out considering $u$ sub.  If you don't see any obvious derivatives, move on to the next method.

anonymous · Answer

hehe well guessing is not working for me...I tried to do it by parts when I first got it and a page and half later I gave up

anonymous · Answer

When you see those forms I just wrote a moment ago, you know immediately you need trig sub.

anonymous · Answer

Trig subs stick out like a sore thumb.

anonymous · Answer

kk I'll try to remember

anonymous · Answer

Parts... it not an easy method to master.  In general you are looking for which part has an nice integral and which part has a nice derivative.

anonymous · Answer

When it comes to rational functions... that is polynomial over polynomial...
Then:
1. Try to factor
2. Trig sub?
3. U sub

If those don't work, you're kinda screwed.

anonymous · Answer

If something not working was an objective standard lol
thanks for help I'll let you go help someone a little less incorrigible