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anonymous
 4 years ago
Integrate (x^2  2x 1) / ( (x1)^2 * (x^2 +1) ).
anonymous
 4 years ago
Integrate (x^2  2x 1) / ( (x1)^2 * (x^2 +1) ).

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

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

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

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

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

TuringTest
 4 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
 4 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
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0I'll be right back in 6 minutes.

TuringTest
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0If you could write it out, i'd really appreciate it.

TuringTest
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0but i understand the rest of what you wrote.

TuringTest
 4 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
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok :). Thank you so much!

TuringTest
 4 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.)
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