# Integral using partial fractions

I am having trouble using partial fractions to evaluate $$\int \frac{6x}{x^3-8} dx.$$ I can find the denominator but using equations to find the numerator is difficult.

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yes, I'm sorry. I'm new to the format here. – Marc D. Feb 28 at 21:22

Partial fractions for: $\frac{6x}{x^3-8}$

$$\frac{6x}{(x-2)(x^2+2x+4)}=\frac{C+Bx}{x^2+2x+4}+\frac{A}{x-2}$$

$$6x=A(x^2+2x+4)+(x-2)(C+Bx)$$

$$6x=4A-2C+(A+B)x^2+(2A-2B+C)x$$

$$\begin{cases} 4A-2C=0\\ 2A-2B+C=6\\ A+B=0 \end{cases}$$

Solve

$$\int \frac{6x}{x^3-8}dx$$

$$=6\int\frac{dx}{(x-2)(x^2+2x+4)}\overset{\text{partial fractios}}{=}\int\frac{2-x}{x^2+2x+4}dx+\int\frac{dx}{x-2}$$

$$=-\frac 1 2\int\frac{2x+2}{x^2+2x+4}dx+3\int\frac{dx}{x^2+2x+4}+\int\frac{dx}{x-2}$$

Set $t=x^2+2x+4$ and $dt=(2x+2)dx$

$$=-\frac 1 2 \int \frac 1 t+3\int\frac{dx}{x^2+2x+4}+\ln|x-2|$$

$$\overset{\text{complete square}}{=}-\frac{\ln|t|}{2}+\ln|x-2|+3\int\frac{dx}{(x+1)^2+3}$$

$$=\color{red}{-\frac{\ln|x^2+2x+4|}{2}+\ln|x-2|+\sqrt 3\arctan\left(\frac{x+1}{\sqrt 3}\right)+\mathcal C}$$

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Thank you for the full answer! – Marc D. Feb 28 at 22:42
@MarcD. You're welcome – Nehorai Feb 28 at 22:44

We want to perform partial fraction of $\frac{6x}{x^3-8},$ therefore $$\frac{6x}{x^3-8} = \frac{6x}{(x-2)(x^2+2x+4)} = \frac{Ax+B}{x^2+2x+4} + \frac{C}{x-2},$$ hence $$\frac{6x}{(x-2)(x^2+2x+4)} = \frac{Ax^2-2Ax+Bx-2B+Cx^2+2Cx+4C}{(x-2)(x^2+2x+4)} = \frac{(A+C)x^2+(-2A+B+2C)x+(4C-2B)}{(x-2)(x^2+2x+4)},$$ and here is our system $$\begin{cases} A+C = 0\\ 2C+B-2A = 6\\ 4C-2B=0 \end{cases},$$ solving the system we obtain $$\begin{cases} C = 1\\ B = 2\\ A=-1 \end{cases},$$

can you do the integration, having $\frac{6x}{x^3-8} = \frac{2-x}{x^2+2x+4} + \frac{1}{x-2}?$

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Yes, thank you I will be able to integrate after the partial fraction expansion. – Marc D. Feb 28 at 21:46
Im so new that I cannot yet upvote ;) – Marc D. Feb 28 at 22:39

First, factor $x^3-8$. $x^3-8=(x^2+2x+4)(x-2)$ $$\frac{6x}{x^3-8}=\frac{ax+b}{x^2+2x+4}+\frac{c}{x-2}$$ Then multiply the lhs denominator out on the right side: $$6x=(x-2)(ax+b)+c(x^2+2x+4)$$ When $x=2$, $(x-2)(ax+b)$ vanishes and $$6(2)=c(4+4+4)\Rightarrow c=1$$ Now, we can use $c$ to solve for $ax+b$. $$6x-(1)(x^2+2x+4)=(x-2)(ax+b)$$ which gives us $a=-1$ and $b=2$. Thus $$\int\frac{6x}{x^3-8}dx=\int\frac{2-x}{x^2+2x+4}+\frac{1}{x-2}dx$$

Which gives you what echzhen just demonstrated.

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This method is the one closest to the way my professor teaches it, while echzhens meathod is similar to the khan academy one which is also very helpful. Thank you for your responses. – Marc D. Feb 28 at 22:39
I think you must mean C=1 – Marc D. Feb 28 at 22:48
Yes, good catch – Matt G Feb 28 at 22:49
You also want $6x=(x−2)(ax+b)+\mathbf c(x^2+2x+4)$ – Paul Evans Feb 29 at 1:23