Question

any body know how to integrate (1)/(x^3+3x^2) using partial fractions?

Answer 1

First step, factor out the bottom. You'll get \(x^2(x+3)\). Then, set it up like this: \[
\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}
\]Set that equal to your equation, and multiply everything by the denominator.\[
1=A(x^2+3x)+B(x+3)+C(x^2)\\
1=(A+C)x^2+(3A+B)x+3B
\]Now you have the following system of equations:\[
A+C=0\\
3A+B=0\\
3B=1
\]This means that:\[
B=\frac{1}{3}, A=\frac{-1}{9}, C=\frac{1}{9}
\]So, the equation can be rewritten as:\[
\frac{\frac{-1}{9}}{x}+\frac{\frac{1}{3}}{x^2}+\frac{\frac{1}{9}}{x+3}
\]You can test that this is right by recombining the fractions to see if it adds back up correctly. Now, integrate each term individually:\[
\frac{-1}{9}\int\frac{1}{x}dx+\frac{1}{3}\int\frac{1}{x^2}dx+\frac{1}{9}\int\frac{1}{x+3}dx
\]

Answer 2

shouldn't it be A(x^2)(x+3)+B(x)(x+3)+C(x)(x^3)?

Answer 3

Nope, that's a common misconception. You don't multiply by each of the other denominators. You multiply the whole thing by the original denominator.

Answer 4

sorry. still not understanding that part

Answer 5

You don't need to multiply A by \(x^2(x+3)\), because A is already over an x, so you just need \(x(x+3)\) to reach the common denominator.

Answer 6

\[
\frac{1}{x^2(x+3)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}\\
\frac{x^2(x+3)}{x^2(x+3)}=\frac{Ax^2(x+3)}{x}+\frac{Bx^2(x+3)}{x^2}+\frac{Cx^2(x+3)}{x+3}\\
1=Ax(x+3)+B(x+3)+C(x^2)
\]

Answer 7

I would use Heavy side cover up for B and C and for A just one substitution.

Answer 8

thanks guys. i got it. appreciate it!

Answer 9

since we know what B and C is, how would find A? I set x=-3 to find C. To find B, i set x=0.

Answer 10

My method is just to treat it as a system of equations, as above, and solve the system.

Answer 11

this is a different question. how would you separate 4x+16/(x(x-2)^2) into the A, B, and C form?

Answer 12

Please close this question and open a new one to ask a new question :)