Partial fractions pronblem help

[rapple]

Homework Statement

F(X)=$$\int[/\frac{1}{1+t^3}<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I have tried different substitutions to find fog where g(t) = ? But am getting stuck$$

 

[Dick]

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

 

[rapple]

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.

 

[Dick]

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.

 

[rapple]

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

How do I proceed

 

[Dick]

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?

 

[rapple]

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

 

[rapple]

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

 

[Dick]

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.

 

[Dick]

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

Ok, you got it. Good work.