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