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January 30, 2015

January 30, 2015

Posted by **Jenna** on Sunday, December 6, 2009 at 9:59pm.

(integral of) (x^2)dx/(x-1)(x^2 +2x+1)

My work:

The above integral is equal to x^2dx/(x+1)^2

(A/x-1) + (B/x+1) + (Cx+D)/(x+1)^2 = x^2

A(x+1)^2 + B(x-1)(x+1) + (Cx+d)(x-1) = x^2

Ax^2 + 2Ax + A + Bx^2 + Cx^2 - Cx -D = x^2

x^2(A+B+C) + x(2A -C+D) + (A-B-D) = x^2

Which gives us the following equations:

A+B+C = 1

2A-C+d= 0

A-B-D=0

The problem is once I get to this point, I get stuck. I'm not sure how to solve for A,B,C, and D.

- calc II -
**MathMate**, Monday, December 7, 2009 at 8:51amYou do not really have a problem.

There are two ways to do a partial fraction with multiple factors in the denominator, and you have combined the two.

For the multiple factor of (x+1)², you could*either*assume

B/(x+1)+D/(x+1)²*or*

(Cx+D)/(x+1)²

In your particular problem, this is the equivalent of setting C=0, and proceed to solve for A,B and D with the three equations that you have correctly set up.

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