Integrate (x^2 - 2x -1) / ( (x-1)^2 * (x^2 +1) ).

- anonymous

Integrate (x^2 - 2x -1) / ( (x-1)^2 * (x^2 +1) ).

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- TuringTest

factor the top and see what happens

- anonymous

ok, let me try that...

- anonymous

I dont think the top can be reduced any further

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- TuringTest

oh I read a plus... my mistake

- anonymous

Partial fraction not helping?

- TuringTest

yeah, I was trying to avoid it
I completed the square on the top which let me split the integral into
1/(x^2+1)-2/[(x-1)^2(x^2+1)]
I suppose you could do partial fractions on the second part, but it will be annoying

- anonymous

I'm not sure how to begin splitting it for partial fractions. I'm wondering if I should write it as A/(x-1)^2 + B/(x^2 + 1) or A/(x-1) + B/(x-1) + C/(x^2 + 1).

- anonymous

ok

- TuringTest

I am trying
A/(x-1) + B/(x-1)^2 + (Cx+D)/(x^2 + 1)
you need the x on top for the squared x in the denom

- TuringTest

do you see how I split the integral up in the first place by completing the square on top?

- anonymous

yes

- TuringTest

so then it's just this ugly PF thing...
I'm gonna collect terms and give it to wolf to solve I think

- anonymous

ok

- anonymous

I'll be right back in 6 minutes.

- TuringTest

ok I got
A=1
B=-1
C=-1
D=0
so we should be able to do this now

- anonymous

ok

- TuringTest

yes, that works, I checked it with wolf.
Want me to write out the whole thing or are you good?

- anonymous

If you could write it out, i'd really appreciate it.

- TuringTest

ok, but I'm not going to show all the partial fractions stuff...
here we go:

- anonymous

i'm still getting use to PF, so if u can, that would be great.

- anonymous

but i understand the rest of what you wrote.

- TuringTest

there are a few ways to do it, but here's the way I did it...

- TuringTest

oh that was great everything I did just got erased. let's try that again...

- TuringTest

\[\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}dx=\int\frac{(x-1)-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}dx\]just looking at the PF part:\[\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}=\frac{-2}{(x-1)^2(x^2+1)}\]multiply out by the denominator on the RHS to get rid of the fractions we obtain\[A(x-1)(x^2+1)+B(x-1)^2+(Cx+D)(x-1)^2=-2\]distribute the parentheses:\[A(x^3-x^2+x-1)+B(x^2+1)+C(x^3-2x^2+x)+D(x^2-2x+1)\]collect all like terms:\[x^3(A+C)+x^2(-A+B-2C+D)+x(A+C-2D)+(-A+B+D)\]now because this all equals -2 we know that the all the terms with x must have a coefficient of 0, so this leads us to the system\[x^3:A+C=0\]\[x^2:-A+B-2C+D=0\]\[x:A+C-2D=0\]\[-A+B+D=-2\]we can solve this in many ways, but here I will just give it to wolfram:http://www.wolframalpha.com/input/?i=A%2BC%3D0%2C-A%2BB-2C%2BD%3D0%2CA%2BC-2D%3D0%2C-A%2BB%2BD%3D-2
so now we know that\[A=1,B=-1,C=-1,D=0\]which leads to\[=\int\frac{1}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}-\frac{x}{x^2+1}dx\]\[=\tan^{-1}x+\ln|x-1|+\frac{1}{x-1}+\frac{1}{2}\ln|x^2+1|+C\]

- TuringTest

whew! that took a while.
I bet there is a faster way to solve the system. For instance letting x=1 in the formula on the 5th line leads directly to\[2B=-2\to B=-1\]from which we can probably solve the system through substitution, but that was giving me a headache so I gave up.
Better to use techniques from linear algebra like Gaussian elimination, or Cramer's rule, but you may not know those techniques yet.

- anonymous

ok :). Thank you so much!

- TuringTest

welcome :)

- TuringTest

\[\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}dx=\int\frac{(x-1)-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}dx\]just looking at the PF part:\[\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}=\frac{-2}{(x-1)^2(x^2+1)}\]multiply out by the denominator on the RHS to get rid of the fractions we obtain\[A(x-1)(x^2+1)+B(x^2+1)+(Cx+D)(x-1)^2=-2\]distribute the parentheses:\[A(x^3-x^2+x-1)+B(x^2+1)+C(x^3-2x^2+x)+D(x^2-2x+1)\]collect all like terms:\[x^3(A+C)+x^2(-A+B-2C+D)+x(A+C-2D)+(-A+B+D)\]now because this all equals -2 we know that the all the terms with x must have a coefficient of 0, so this leads us to the system\[x^3:A+C=0\]\[x^2:-A+B-2C+D=0\]\[x:A+C-2D=0\]\[-A+B+D=-2\]we can solve this in many ways, but here I will just give it to wolfram:http://www.wolframalpha.com/input/?i=A%2BC%3D0%2C-A%2BB-2C%2BD%3D0%2CA%2BC-2D%3D0%2C-A%2BB%2BD%3D-2
so now we know that\[A=1,B=-1,C=-1,D=0\]which leads to\[=\int\frac{1}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}-\frac{x}{x^2+1}dx\]\[=\tan^{-1}x+\ln|x-1|+\frac{1}{x-1}+\frac{1}{2}\ln|x^2+1|+C\]
(there was a small typo above the last time, so I fixed it so as not to confuse you.)

Looking for something else?

Not the answer you are looking for? Search for more explanations.