Integral of 1/(t^4-25) using trig substitution

[ziddy83]

ok here is the integral...sorry for the laziness...

integral of dt/(t^4)-25

would this just turn out to be (t^4)-25 since you can bring up the denominator to the top as (something)^-1...then just take the antiderivative...? hope that makes sense

edit: ok that won't work...so I think i need to make a trig sub...so like...hmm do i let t= sec x?

 

[arildno]

Rewrite it to:
$$I=\int\frac{dt}{(t^{2}+5)(t^{2}-5)}$$
and use partial fractions decomposition.

 

[whozum]

$$\int \frac{dt}{t^4-25} = \int \frac{dt}{(t^2-5)(t^2+5)}$$

edit: tex error

 

[ziddy83]

oh ok...cool. Thanks guys

 

[GCT]

Try factoring out $$t^{4}-25$$. Didn't you learn how to do these type of integrals in lecture (partial fractions)? You'll have to learn how to do such integrals yourself, the process is quite tedious.

 

[arildno]

That was quick, GCT..

 

[GCT]

it seems that we were all answering this question at the same time.