# Integral Exercise!

• March 14th 2009, 01:18 PM
ypatia
Integral Exercise!
Find the function f where f(x) is f(x)=∫((x-1)^2)/(x^2+1)) f(x)dx - c
• March 14th 2009, 01:21 PM
TheEmptySet
Quote:

Originally Posted by ypatia
Find the function f where f(x) is f(x)=∫((x-1)^2)/(x^2+1)) f(x)dx - c

$\frac{(x-1)^2}{x^2+1}=\frac{x^2-2x+1}{x^2+1}=\frac{(x^2+1)-2x}{x^2+1}=1-\frac{2x}{x^2+1}$

This is alot better to integrate.

I hope this helps
• March 29th 2009, 12:52 AM
The Second Solution
Quote:

Originally Posted by ypatia
Find the function f where f(x) is f(x)=∫((x-1)^2)/(x^2+1)) f(x)dx - c

Differentiate both sides of the given equation:

$f'(x) = \frac{(x-1)^2}{x^2+1} f(x)$

Now re-arrange and integrate: $\int \frac{f'(x)}{f(x)} \, dx = \int \frac{(x-1)^2}{x^2+1} \, dx$.

The integral on the right hand side is found using the suggestion in the previous reply.