5x

(x-1)((x^2)+4)

fractions?

5x =A+B

(x-1) ?

And ((x^2)+4) is 1/2arctan(x/2)

the answer key says

ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)

Can anyone help? Thanks.

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- Jun 4th 2008, 02:48 PMkhuezyHow would you integrate this?
__5x__

(x-1)((x^2)+4)

fractions?

5x =__A__+__B__

(x-1) ?

And ((x^2)+4) is 1/2arctan(x/2)

the answer key says

ln|x-1| -.5ln|x^2+4| - 2arctan(x/2)

Can anyone help? Thanks. - Jun 4th 2008, 03:06 PMo_O
(since is an irreducible quadratic, indicating your numerator is in the form of Bx + C)

See what you can do with this and come back with any questions. - Jun 4th 2008, 03:08 PMTheEmptySet
You can use partial fractions with this setup

Or observe that

Now we have

The first is in the form of the arctangent or if you want to do it by hand let x=2tan(u) the 2nd is a u sub on the denominator and the last one is a u sub on the denominator .

I hope this helps.

Good luck. - Jun 4th 2008, 03:33 PMkhuezyhelp
I got

5x = (A+B)x^2 + 4A - (B+C)x - C

I'm confused on what to do next.

Equating coefficients?

A+B = 0 4A=5 B+C=0 C=0????????????????

Thanks. - Jun 4th 2008, 03:39 PMgalactus
Here's another way using trig.

Let

Making the subs we get a menacing looking thing, but it simplifies down fairly nice.

Making the subs we get:

This, believe it or not, whittles down to

Then, after integrating, resub in

You may wish to stick with the others methods. I just threw it out there. - Jun 4th 2008, 04:03 PMMathstud28