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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

factor the top and see what happens
ok, let me try that...
I dont think the top can be reduced any further

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

oh I read a plus... my mistake
Partial fraction not helping?
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
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).
ok
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
do you see how I split the integral up in the first place by completing the square on top?
yes
so then it's just this ugly PF thing... I'm gonna collect terms and give it to wolf to solve I think
ok
I'll be right back in 6 minutes.
ok I got A=1 B=-1 C=-1 D=0 so we should be able to do this now
ok
yes, that works, I checked it with wolf. Want me to write out the whole thing or are you good?
If you could write it out, i'd really appreciate it.
ok, but I'm not going to show all the partial fractions stuff... here we go:
i'm still getting use to PF, so if u can, that would be great.
but i understand the rest of what you wrote.
there are a few ways to do it, but here's the way I did it...
oh that was great everything I did just got erased. let's try that again...
\[\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}dx=\int\frac{(x-1)-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}dx\]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\[=\int\frac{1}{x^2+1}+\frac{1}{x-1}-\frac{1}{(x-1)^2}-\frac{x}{x^2+1}dx\]\[=\tan^{-1}x+\ln|x-1|+\frac{1}{x-1}+\frac{1}{2}\ln|x^2+1|+C\]
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.
ok :). Thank you so much!
welcome :)
\[\int\frac{x^2-2x-1}{(x-1)^2(x^2+1)}dx=\int\frac{(x-1)-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{-2}{(x-1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+1}dx\]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.)

Not the answer you are looking for?

Search for more explanations.

Ask your own question