Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions

Question

I'm having issues with the Partial Fractions method:

$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$

I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.

So, first of all, we must factorize the denominator:

$$x^3+2x^2 = (x+2)\cdot x^2$$

Great. So now we write three fractions:

$$\frac{A}{x^2} + \frac{B}{x} + \frac{C}{x+2}$$

Eventually we conclude that

$$A(x+2)+B(x+2)(x)+C(x^2) = 5x^2+3x-2$$

So now we look at what happens when $x = -2$:

$$C = 12$$

When $x = 0$:

$$A = -1$$

And now we are missing $B$, but we can just pick an arbitrary number for $x$ like... $1$:

$$B = -1$$

We replace the values here:

$$\int \frac{-1}{x^2}dx + \int \frac{-1}{x}dx + \int \frac{12}{x+2}dx$$

Which results in

$$\frac{1}{x}-\ln(x)+12\ln(x+2)+K$$

But I fear the answer actually is

$$\frac{1}{x}+2\ln(x)+3\ln(x+2)+K$$

Can you tell me what did I do wrong, and what should I have done?

Answer 1

You have an algebra error. When you put $x = -2$, the constraint equation collapses to $4C = 12$, so $C = 3$.

Your method for determining $B$ is correct – the constraining identity must hold for any $x$. Putting $x = 1$ is a fine choice as the arithmetic is easier. However the incorrect value for $C$ will affect the value you determine for $B$.

Answer 2

You got $A$ correct.

You have incorrectly evaluated $C$. When you sub in $x=-2$ you should get:

$$5(-2)^2+3(-2)-2=(-2)^2C$$

$$12=4C$$

$$C=3$$

Then to evaluate $B$ you can pick an arbitrary value for $x$ then solve for $B$. E.g. you selected $x=1$ this gives:

$$5\cdot1^2+3\cdot1-2=A\cdot(1+2)+B\cdot(1+2)\cdot1+C\cdot1^2$$

Replace $A$ and $C$ with the values you already found to give:

$$5+3-2=-3+3B+3$$

$$6=3B$$

$$B=2$$

Answer 3

Once you arrive at $$A(x+2) + B(x+2)x + C(x^2) = 5x^2 + 3x - 2,$$ this equation must hold true for all $x$. If you substitute $x = -2$, then $$C(-2)^2 = 5(-2)^2 + 3(-2) - 2 = 5(4) - 6 - 2 = 12,$$ or $4C = 12$, or $$C = 3.$$ You seem to be under the impression that you can disregard any expressions involving $x$ on the left-hand side. That is incorrect.

Next, substituting $x = 0$ gives $$A(0+2) = 5(0^2) + 3(0) - 2 = -2,$$ or $2A = -2$, or $$A = -1.$$ This you got right. Now finally, if you were to choose $x = 1$, you would obtain $$A(1+2) + B(1+2)(1) + C(1^2) = 5(1^2) + 3(1) - 2 = 6,$$ and the left-hand side simplifies to $3A + 3B + C$. To get $B$, we now have to substitute the values you previously obtained for $A$ and $C$: $$3(-1) + 3B + 3 = 6,$$ or $$B = 2.$$ Therefore, the solution is $$\frac{5x^2 + 3x - 2}{x^2(x+2)} = \frac{-1}{x^2} + \frac{2}{x} + \frac{3}{x+2}.$$

Answer 4

The value of $A$ is right. $A = -1$. When you plug in $x = -2$, you've forgotten to divide it by $x^2$, so you get $C = 12$. Actually, $C = 3$.

\begin{align} -(x+2) + Bx(x+2) + 3x^2 &= 5x^2+3x-2 \\ 3x^2-x-2 + Bx(x+2) &= 5x^2+3x-2 \\ Bx(x+2) &= 2x^2 + 4x = 2x(x+2)\\ B &= 2 \end{align}

\begin{align} & \int \frac{-1}{x^2}dx + \int \frac{2}{x}dx + \int \frac{3}{x+2}dx \\ =& \frac{1}{x}+2\ln(x)+3\ln(x+2)+K \end{align}

Answer 5

I can't resist to post this. I think this is the easiest way.

$$(B+C)x^2+(A+2B)x+2A=5x^2+3x-2$$ \begin{align} 2A&=-2&\rightarrow A&=-1\\ A+2B&=-1+2B=3&\rightarrow B&=2\\ B+C&=2+C=5&\rightarrow C&=3 \end{align}

Answer 6

$$(3X^2+4X+2X^2-X-2)/(X^3+2X^2 )=(3X^2+4X)/(X^3+2X^2 )+(2X+3)/X^2 -8/(X+2)$$ $ln(X^3+2X^2 )+lnX^2-3/x-8ln(X+2)+c$