# Integrate problem with partial fractions

• August 27th 2009, 04:45 AM
Karl Harder
Integrate problem with partial fractions
Evaluate

$\int_{}^{t}dt\frac{t^3+a^3}{t^3-a^3}$

So far i have

$\int_{}^{t}dt\frac{t^3+a^3}{t^3-a^3} = \int_{}^{t}dt\frac{t^3-a^3+a^3+a^3}{t^3-a^3} = \int_{}^{t}dt\frac{t^3-a^3}{t^3-a^3}+\frac{2a^3}{t^3-a^3} = \int_{}^{t}dt1+\frac{2a^3}{t^3-a^3}$

I'm not sure where i am to go next. Any hints as to what i can do?
• August 27th 2009, 07:34 AM
JG89
$\int 1 + \frac{2a^3}{x^3 - a^3} dt = t + 2a^3 \int \frac{dt}{x^3 - a^3}$

Use the "difference of cubes" formula and the denominator should factor to a product of a linear expression and an irreducible quadratic. Then you can use partial fractions.
• August 27th 2009, 07:41 AM
VonNemo19
Quote:

Originally Posted by Karl Harder
Any hints as to what i can do?

Factor...