Question

Integral of x^2/(x^3+1)
which is the general solution to start off with?

Answer 1

Let x^3+1=u
Differentiate both sides to get :
3x^2 = du/dx
dx=du/3x^2

Now you have to integrate simply\[\Huge \int\limits \frac{x^2 }{u } \times \frac{du}{3x^2}\]

Answer 2

I started off with u = x^2 and du = 2x
dv = 1/x^3+1 that gives v = kn (x^3+1)
then u'v+uv'

Answer 3

nope you can't let u=x^2

Answer 4

oh okey thanks, I'll try to solve it.

Answer 5

:)

Answer 6

okey so I shorten it down first to:
\[\frac{ 1 }{ 3 } \int\limits_{}^{} \frac{ x^2 }{ x^3+1 } * \frac{ 3x^2 }{ 3x^2 }\]
then the remaining equation is:
\[\frac{ 1 }{ 3 } \int\limits_{}^{} \frac{ x^2 }{ x^3 +1}\]
Where do I go from here?

Answer 7

@Kainui

Answer 8

Okey first misstake realised is the 1/3, it should be remoived.

Answer 9

\[\frac{ x^3 }{ 3 } \ln x^3+1 +c\] now how does that convert into
\[\frac{ 1 }{ 3 } \ln x^3 +1+ c\]

Answer 10

Hmm this seems fishy to me, let's start over.
\[\Large \int\limits \frac{x^2}{x^3+1}dx\]

Now we should notice that the derivative of x^3 is really just x^2 times a constant. That is usually a good sign that we should pick x^3 to be used for u-substitution. Since the derivative of a constant is 0, we also pick the 1 along with it

\[\Large u=x^3+1\]
Now we take the derivative \[\Large \frac{du}{dx}=3x^2\]and then we rearrange it so it looks like what we have left so that we can plug it into the dx part of the integral with the x^2.\[\Large \frac{du}{3}=x^2dx\]See? Just algebra.

Ok, so if we plug this in, I'll sort of show you in steps to see how I'm doing this: \[\Large \int\limits \frac{x^2}{x^3+1}dx=\int\limits (\frac{1}{x^3+1})*(x^2dx)\] Now we substitute thes things both in: \[\Large \int\limits (u)*(\frac{du}{3})=\frac{1}{3} \int\limits u du\]Now you can siply integrate this and then take your original u from earlier and plug it into the answer to get it back in terms of x.

Answer 11

Whoops I think I messed up right at the end there haha! \[\Large \int\limits (\frac{1}{x^3+1})*(x^2dx)=\int\limits (\frac{1}{u})*(\frac{du}{3})\]

Answer 12

okey I solved it, with your help, sometimes I make it harder than it needs to be.
\[\frac{ 1 }{ 3 } \ln x^3+1 +c\]

Answer 13

\[\Large \int\frac{1}{u}du=\ln|u|+c\]

Answer 14

yes, I made the misstake by putting in the numbers in du and u and started dividing and so on. @Kainui your explanation was very clear, thanks.

Answer 15

Thanks to Leibniz, infinitesimals do work like algebra in operations like these. I think it is a stretch to say these operations are algebra.

Answer 16

I think it's a stretch to say it's not algebra.

Answer 17

http://openstudy.com/study#/updates/53c27f8de4b09e08a1c7c2f1

Answer 18

Okey I'll close this now, and continue with a simular integral and hopefully be able to solve,
thanx to evry1 for taking their time!:)