Partial fractions pronblem help

In summary, the conversation involved finding the derivative of the integral of a function with the upper limit as a function of x. The solution involved using the Fundamental Theorem of Calculus and the Leibniz integral rule to find the derivative, which is equal to the function itself multiplied by the derivative of the upper limit. After some trial and error, the correct solution was found to be the function itself multiplied by 2x.
  • #1
rapple
25
0

Homework Statement


F(X)=[tex]\int[/\frac{1}{1+t^3}


Homework Equations





The Attempt at a Solution


I have tried different substitutions to find fog where g(t) = ? But am getting stuck
 
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  • #2


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


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


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


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


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


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


Ok.
F'(x)=d/dx(integ f(t)) over 0 to x^2. = f(x).2x=(1/1+x^6).2x
 
  • #9


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


rapple said:
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.
 

FAQ: Partial fractions pronblem help

What are partial fractions and why are they important?

Partial fractions are a method for breaking down a rational function into simpler fractions. They are important because they allow us to solve integrals and differential equations more easily, and they also help us understand the behavior of functions.

How do I know when to use partial fractions?

You should use partial fractions when you have a rational function with a denominator that can be factored into linear and/or irreducible quadratic terms. This means that the denominator cannot be further simplified and the numerator is of lower degree than the denominator.

What are the steps for solving a partial fractions problem?

The general steps for solving a partial fractions problem are: 1) Factor the denominator of the rational function, 2) Write the partial fractions with undetermined coefficients, 3) Find the values of the coefficients by equating the original function to the partial fractions, 4) Integrate both sides of the equation, and 5) Solve for any remaining variables or constants.

Can I use partial fractions for improper rational functions?

Yes, you can use partial fractions for improper rational functions as long as the degree of the numerator is less than the degree of the denominator. If the degree of the numerator is equal to or greater than the degree of the denominator, you will need to use polynomial division to simplify the function before applying partial fractions.

Are there any common mistakes to avoid when solving partial fractions?

Yes, some common mistakes to avoid when solving partial fractions are: 1) Forgetting to factor the denominator, 2) Using incorrect values for the coefficients, 3) Missing terms when equating the original function to the partial fractions, and 4) Making algebraic errors when integrating or solving for variables. It is important to double check your work and simplify your final answer to ensure it is correct.

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