Asked by Jenna
Express the integrals as the sum of partial fractions and evaluate the integral:
(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.
(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.
Answers
Answered by
MathMate
You 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 <i>either</i> assume
B/(x+1)+D/(x+1)² <i>or</i>
(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.
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 <i>either</i> assume
B/(x+1)+D/(x+1)² <i>or</i>
(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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