what is antideriv. of arctan?

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January 30th 2008, 03:18 PM
Tuugii

what is antideriv. of arctan?

hey,

I've got this problem:

[indefinite integral](1/(t^3+t))dt

I've used the partial fractions to split the integral into 2 separate parts, which becomes:

ln|t| + [indefinite integral](t/(t^2 + 1))dt + C

here I used the integration by parts for the bolded integral, which gives me:

ln|t| + t*arctan(t) + [definite integral](arctan(t))dt + C

and I cannot figure out what is the antiderivative of arctan(t) is?

thanks for the tips,
Tuugii
January 30th 2008, 03:28 PM
polymerase

$\int \frac{1}{t}-\frac{t}{t^2+1}=\int \frac{1}{t}-\frac{1}{2}\int \frac{(t^2+1)'}{t^2+1}=ln|t|-\frac{1}{2}ln|t^2+1|$

no arctan involved just regualr subtitution:D
January 30th 2008, 03:33 PM
Pinsky

i don't know how to solve arctg, but why didn't u use a subtitution
$t^2+1=u$
That would make the integral solvable.
January 30th 2008, 03:47 PM
Jhevon

As the previous posts say, dealing with arctangent is not an issue here. however, if you are still curious, we would find that integral using integration by parts, with $u = \arctan x$ and $dv = 1$
January 30th 2008, 04:08 PM
polymerase

Quote:

Originally Posted by Tuugii

I've used the partial fractions to split the integral into 2 separate parts,

you dont even need to use that for this question, check this out

$\int \frac{1}{t(t^2+1)}= \frac{1+t^2-t^2}{t(t^2+1)}= \int \frac{1+t^2}{t(t^2+1)}-\frac{t^2}{t(t^2+1)}=\int \frac{1}{t}-\frac{t}{t^2+1}$
January 30th 2008, 04:52 PM
Tuugii

thanks a lot for the replies. :)

was really helpful! :)

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