A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Integrate (x^2  2x 1) / ( (x1)^2 * (x^2 +1) ).
anonymous
 5 years ago
Integrate (x^2  2x 1) / ( (x1)^2 * (x^2 +1) ).

This Question is Closed

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2factor the top and see what happens

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, let me try that...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont think the top can be reduced any further

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2oh I read a plus... my mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Partial fraction not helping?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2yeah, 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/[(x1)^2(x^2+1)] I suppose you could do partial fractions on the second part, but it will be annoying

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure how to begin splitting it for partial fractions. I'm wondering if I should write it as A/(x1)^2 + B/(x^2 + 1) or A/(x1) + B/(x1) + C/(x^2 + 1).

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2I am trying A/(x1) + B/(x1)^2 + (Cx+D)/(x^2 + 1) you need the x on top for the squared x in the denom

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2do you see how I split the integral up in the first place by completing the square on top?

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2so then it's just this ugly PF thing... I'm gonna collect terms and give it to wolf to solve I think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll be right back in 6 minutes.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2ok I got A=1 B=1 C=1 D=0 so we should be able to do this now

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2yes, that works, I checked it with wolf. Want me to write out the whole thing or are you good?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you could write it out, i'd really appreciate it.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2ok, but I'm not going to show all the partial fractions stuff... here we go:

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm still getting use to PF, so if u can, that would be great.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i understand the rest of what you wrote.

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2there are a few ways to do it, but here's the way I did it...

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2oh that was great everything I did just got erased. let's try that again...

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2\[\int\frac{x^22x1}{(x1)^2(x^2+1)}dx=\int\frac{(x1)2}{(x1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{2}{(x1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x1}+\frac{B}{(x1)^2}+\frac{Cx+D}{x^2+1}dx\]just looking at the PF part:\[\frac{A}{x1}+\frac{B}{(x1)^2}+\frac{Cx+D}{x^2+1}=\frac{2}{(x1)^2(x^2+1)}\]multiply out by the denominator on the RHS to get rid of the fractions we obtain\[A(x1)(x^2+1)+B(x1)^2+(Cx+D)(x1)^2=2\]distribute the parentheses:\[A(x^3x^2+x1)+B(x^2+1)+C(x^32x^2+x)+D(x^22x+1)\]collect all like terms:\[x^3(A+C)+x^2(A+B2C+D)+x(A+C2D)+(A+B+D)\]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\[x^3:A+C=0\]\[x^2:A+B2C+D=0\]\[x:A+C2D=0\]\[A+B+D=2\]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%2CA%2BB2C%2BD%3D0%2CA%2BC2D%3D0%2CA%2BB%2BD%3D2 so now we know that\[A=1,B=1,C=1,D=0\]which leads to\[=\int\frac{1}{x^2+1}+\frac{1}{x1}\frac{1}{(x1)^2}\frac{x}{x^2+1}dx\]\[=\tan^{1}x+\lnx1+\frac{1}{x1}+\frac{1}{2}\lnx^2+1+C\]

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2whew! 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok :). Thank you so much!

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.2\[\int\frac{x^22x1}{(x1)^2(x^2+1)}dx=\int\frac{(x1)2}{(x1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{2}{(x1)^2(x^2+1)}dx\]\[=\int\frac{1}{x^2+1}+\frac{A}{x1}+\frac{B}{(x1)^2}+\frac{Cx+D}{x^2+1}dx\]just looking at the PF part:\[\frac{A}{x1}+\frac{B}{(x1)^2}+\frac{Cx+D}{x^2+1}=\frac{2}{(x1)^2(x^2+1)}\]multiply out by the denominator on the RHS to get rid of the fractions we obtain\[A(x1)(x^2+1)+B(x^2+1)+(Cx+D)(x1)^2=2\]distribute the parentheses:\[A(x^3x^2+x1)+B(x^2+1)+C(x^32x^2+x)+D(x^22x+1)\]collect all like terms:\[x^3(A+C)+x^2(A+B2C+D)+x(A+C2D)+(A+B+D)\]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\[x^3:A+C=0\]\[x^2:A+B2C+D=0\]\[x:A+C2D=0\]\[A+B+D=2\]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%2CA%2BB2C%2BD%3D0%2CA%2BC2D%3D0%2CA%2BB%2BD%3D2 so now we know that\[A=1,B=1,C=1,D=0\]which leads to\[=\int\frac{1}{x^2+1}+\frac{1}{x1}\frac{1}{(x1)^2}\frac{x}{x^2+1}dx\]\[=\tan^{1}x+\lnx1+\frac{1}{x1}+\frac{1}{2}\lnx^2+1+C\] (there was a small typo above the last time, so I fixed it so as not to confuse you.)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.