# Partial fractions pronblem help

1. Apr 9, 2009

### rapple

1. The problem statement, all variables and given/known data
F(X)=[tex]\int[/\frac{1}{1+t^3}

2. Relevant equations

3. The attempt at a solution
I have tried different substitutions to find fog where g(t) = ? But am getting stuck

2. Apr 9, 2009

### Dick

Re: Antiderivative

1+t^3 can be factored. Start by using partial fractions.

3. Apr 9, 2009

### rapple

Re: Antiderivative

I tried partial fractions but I landed up with A/(1+t) + B/(1-t+t^2). Cannot find values for A & B that work.

4. Apr 9, 2009

### Dick

Re: Antiderivative

Try A/(1+t)+(B*t+C)/(1-t+t^2). If you have a quadratic in the denominator it's not necessarily a constant in the numerator.

5. Apr 9, 2009

### rapple

Re: Antiderivative

I got the problem wrong.
F(x)=Integ (1+t^3)^-1 from 0 to x^2. Find F'(x)

How do I proceed

6. Apr 9, 2009

### Dick

Re: Antiderivative

That makes your job a lot easier. Look up the Fundamental Theorem of Calculus and the Leibniz integral rule. What's the derivative of an integral?

7. Apr 9, 2009

### rapple

Re: Antiderivative

The derivative of an integral is the function itself if it is continuous over the specified region. In this case, the function is not continuous at t=-1, But that is not in 0 to x^2.

I don't know how Leibniz rule works here

8. Apr 9, 2009

### rapple

Re: Antiderivative

Ok.
F'(x)=d/dx(integ f(t)) over 0 to x^2. = f(x).2x=(1/1+x^6).2x

9. Apr 9, 2009

### Dick

Re: Antiderivative

That's not good. Yes, the integral of a function from 0 to x of f(t) is f(x). If the upper limit is not x but some function of x (like x^2) you have to use the chain rule to find d/dx. That's what the Leibniz thing is about.

10. Apr 9, 2009

### Dick

Re: Antiderivative

Ok, you got it. Good work.