find the integral of 1/(t^3(t^2-1)^1/2))

I tried this a multiple of times by treating u as secu. But somehow, the answer ends up as arctan something.

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- Oct 22nd 2010, 03:37 PMTaurus3indefinite integral
find the integral of 1/(t^3(t^2-1)^1/2))

I tried this a multiple of times by treating u as secu. But somehow, the answer ends up as arctan something. - Oct 22nd 2010, 04:22 PMJhevon
yeah, doing this via that trig sub will not get you arctan in your answer (that is the easier way to do this integral though, and if you do it correctly, your answer should be fine). they got their answer using trig sub twice. so if you insist on getting the answer your text got, proceed thusly (the harder way):

Note that your integral is the same as:

$\displaystyle \displaystyle \int \frac t{t^4 (t^2 - 1)^{1/2}}~dt$ ............multiply by $\displaystyle \displaystyle \frac tt$

Now, let $\displaystyle \displaystyle u^2 = t^2 - 1$, the integral becomes:

$\displaystyle \displaystyle \int \frac 1{(u^2 + 1)^2}~du$

$\displaystyle \displaystyle = \int \frac {u^2 + 1 - u^2}{(u^2 + 1)^2}~du$

$\displaystyle \displaystyle = \int \frac 1{u^2 + 1}~du - \int \frac {u^2}{(u^2 + 1)^2}~du$

The first integral will give you your arctan. For the second integral, use an appropriate trig sub.

You should end up with the desired antiderivative.