Finding the integral of a function

In other words, it changes it to ln(u).In summary, the integral of the given function is simply the natural logarithm of the denominator, ln|x^3+2| + a constant of integration, since the numerator is the derivative of the denominator. Substitution is the easiest technique to use in this case.
  • #1
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Homework Statement


Hi, all I need to do is find the integral of a function, and I am struggling with where to begin with this particular expression.


Homework Equations



[itex]\int[(3x^2)/(x^3+ 2)]dx[/itex]

The Attempt at a Solution



Thats the problem, I am not sure where to start! I am sure it is something obvious I am missing.
Thanks

Homework Statement





Homework Equations





The Attempt at a Solution

 
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  • #2
I assume you mean [itex]\int[(3x^2)/(x^3+ 2)]dx[/itex].

Let [itex]u= x^3+ 2[/itex].
 
  • #3
Thanks for your quick reply. Sorry, that is exactly what I meant, I am not great with laying stuff out online.

Substituting was my first thought, however this question is part of a pre-test I am doing for homework, and there is a section on substituting later on in the paper, so I am guessing I'm not supposed to use that here. Is there another way doing it?
 
  • #4
For equations of this form, where the numerator is the derivative of the denominator, the integral is simply the natural logarithm of the denominator. So in this case it would be ln|x^3+2| + a constant of integration.
 
  • #5
iaing94 said:
Substituting was my first thought, however this question is part of a pre-test I am doing for homework, and there is a section on substituting later on in the paper, so I am guessing I'm not supposed to use that here. Is there another way doing it?
Substitution is probably the easiest technique, and the one that should be tried first. I can't think of any other way to do this problem.
 
  • #6
Jamo1991 said:
For equations of this form, where the numerator is the derivative of the denominator, the integral is simply the natural logarithm of the denominator. So in this case it would be ln|x^3+2| + a constant of integration.

I never knew that, that works every time? Care to explain how?
 
  • #7
Because of "substitution" whether you call it that or not! If your integral is of the form [itex]\int f'(x)/f(x)dx[/itex], let u= f(x), du= f'dx changes the integral to [itex]\int du/u[/itex].
 

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