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
$\displaystyle \frac{5x}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+4}$ (since $\displaystyle x^2 + 4$ 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

$\displaystyle \frac{5x}{(x^2+4)(x-1)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+4}$

Or observe that

$\displaystyle \frac{5x}{(x-1)(x^2+4)}=\frac{5x-x^2-4+x^2+4}{(x-1)(x^2+4)}=\frac{-(x^2-5x+4)}{(x-1)(x^+2)}+\frac{x^2+4}{(x-1)(x^2+4)}=$

$\displaystyle \frac{-(x-4)(x-1)}{(x^2+4)(x-1)}+\frac{x^2+4}{(x-1)(x^2+4)}=\frac{4-x}{x^2+4}+\frac{1}{x-1}=\frac{4}{x^2+4}-\frac{x}{x^2+4}+\frac{1}{x-1}$

Now we have

$\displaystyle \int \left( \frac{4}{x^2+4}-\frac{x}{x^2+4}+\frac{1}{x-1} \right)dx$

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 $\displaystyle u=x^2+4$ and the last one is a u sub on the denominator $\displaystyle u=x-1$.

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.

$\displaystyle \frac{x-1}{x^{2}+4}dx$

Let $\displaystyle x=2tan(t), \;\ dx=2sec^{2}(t)dt$

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

Making the subs we get:

$\displaystyle \frac{2tan(t)-1}{4(tan^{2}(t)+1)}\cdot{2sec^{2}(t)}dt$

This, believe it or not, whittles down to

$\displaystyle tan(t)-\frac{1}{2}$

$\displaystyle \int{tan(t)}dt-\frac{1}{2}\int{dt}$

Then, after integrating, resub in $\displaystyle t=tan^{-1}(\frac{x}{2})$

You may wish to stick with the others methods. I just threw it out there. - Jun 4th 2008, 04:03 PMMathstud28
$\displaystyle \frac{5x}{(x^2+4)(x-1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$

multiplying through by the common denominator we get

$\displaystyle 5x=A(x^2+4)+(Bx+C)(x-1)$

Now letting $\displaystyle x=1$ we get

$\displaystyle 5(1)=A(1^2+4)+(B+C)(0)\Rightarrow{A=1}$

So now for the second factor we need to expand, but dont forget A=1

So we have

$\displaystyle 5x=x^2+4+Bx^2-Bx+Cx-C$

So we see that $\displaystyle B+1=0\Rightarrow{B=-1}$

and also seeing $\displaystyle 4-C=0\Rightarrow{C=4}$

So finally we can see that

$\displaystyle \frac{5x}{(x^2+4)(x-1)}=\frac{1}{x-1}+\frac{4-x}{x^2+4}$

and we see that