Integral using partial fractions

Question

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.

Answer 1

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 $dt=(2x+2)dx$

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

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

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

Answer 2

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 $\frac{6x}{x^3-8} = \frac{2-x}{x^2+2x+4} + \frac{1}{x-2}?$

Answer 3

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.