Integrate problem with partial fractions

**Karl Harder** · August 27th 2009, 04:45 AM

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?

**JG89** · August 27th 2009, 07:34 AM

$\int 1 + \frac{2a^3}{x^3 - a^3} dt = t + 2a^3 \int \frac{dt}{x^3 - a^3}$

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.

**VonNemo19** · August 27th 2009, 07:41 AM

Originally Posted by Karl Harder

Any hints as to what i can do?

Factor...

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