How do you integrate $\frac{6 x}{x^{3} - 8}$ $(6x)/(x^3-8)$ using partial fractions?

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How do you integrate $\frac{6 x}{x^{3} - 8}$ $(6x)/(x^3-8)$ using partial fractions?

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Answer 1 · 2017-01-08T06:49:30

The answer is $= ln (∣ x - 2 ∣) - \frac{1}{2} ln (∣ x^{2} + 2 x + 4 ∣) + \frac{4}{\sqrt{3}} arctan (\frac{x + 1}{\sqrt{3}}) + C$ $=ln(∣x-2∣)-1/2ln(∣x^2+2x+4∣)+4/sqrt3arctan((x+1)/sqrt3)+C$

Explanation:

The denominator is

$x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ $x^3-8=(x-2)(x^2+2x+4)$

Therefore,

$\frac{6 x}{x^{3} - 8} = \frac{6 x}{(x - 2) (x^{2} + 2 x + 4)}$ $(6x)/(x^3-8)=(6x)/((x-2)(x^2+2x+4))$

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

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

So,

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

Let $x = 2$ $x=2$ , $\Rightarrow$ $=>$ , $12 = 12 A$ $12=12A$ , $\Rightarrow$ $=>$ , $A = 1$ $A=1$

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $0 = 4 A - 2 C$ $0=4A-2C$ , $\Rightarrow$ $=>$ , $C = 2$ $C=2$

Coefficients of $x^{2}$ $x^2$ , $\Rightarrow$ $=>$ , $0 = A + B$ $0=A+B$ , $\Rightarrow$ $=>$ , $B = - 1$ $B=-1$

Therefore,

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

So,

$\int \frac{6 x d x}{x^{3} - 8} = \int \frac{d x}{x - 2} - \int \frac{(x - 2) d x}{x^{2} + 2 x + 4}$ $int(6xdx)/(x^3-8)=intdx/(x-2)-int((x-2)dx)/(x^2+2x+4)$

We integrate separately

$\int \frac{d x}{x - 2} = ln (∣ x - 2 ∣)$ $intdx/(x-2)=ln(∣x-2∣)$

$\int \frac{(x - 2) d x}{x^{2} + 2 x + 4} = \int \frac{(x + 2 - 4) d x}{x^{2} + 2 x + 4}$ $int((x-2)dx)/(x^2+2x+4)=int((x+2-4)dx)/(x^2+2x+4)$

$= \int \frac{(x + 2) d x}{x^{2} + 2 x + 4} - \int \frac{4 d x}{x^{2} + 2 x + 4}$ $=int((x+2)dx)/(x^2+2x+4)-int(4dx)/(x^2+2x+4)$

Let $u = x^{2} + 2 x + 4$ $u=x^2+2x+4$ , $\Rightarrow$ $=>$ , $d u = (2 x + 2) d x$ $du=(2x+2)dx$

Therefore,

$\int \frac{(x + 2) d x}{x^{2} + 2 x + 4} = \frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} ln u$ $int((x+2)dx)/(x^2+2x+4)=1/2int(du)/u=1/2lnu$

$= \frac{1}{2} ln (∣ x^{2} + 2 x + 4 ∣)$ $=1/2ln(∣x^2+2x+4∣)$

$x^{2} + 2 x + 4 = x^{2} + 2 x + 1 + 3 = {(x + 1)}^{2} + 3$ $x^2+2x+4=x^2+2x+1+3=(x+1)^2+3$

$\int \frac{4 d x}{x^{2} + 2 x + 4} = 4 \int \frac{d x}{{(x + 1)}^{2} + 3}$ $int(4dx)/(x^2+2x+4)=4int(dx)/((x+1)^2+3)$

$= 4 \int \frac{d x}{3 ({(\frac{x + 1}{\sqrt{3}})}^{2} + 1)}$ $=4int(dx)/(3(((x+1)/sqrt3)^2+1))$

$= \frac{4}{3} \int \frac{d x}{({(\frac{x + 1}{\sqrt{3}})}^{2} + 1)}$ $=4/3int(dx)/((((x+1)/sqrt3)^2+1))$

Let $tan θ = \frac{x + 1}{\sqrt{3}}$ $tantheta=(x+1)/sqrt3$ $\Rightarrow$ $=>$ ${sec}^{2} θ d θ = \frac{d x}{\sqrt{3}}$ $sec^2theta d theta=dx/sqrt3$

Therefore,

$\frac{4}{3} \int \frac{d x}{({(\frac{x + 1}{\sqrt{3}})}^{2} + 1)} = \frac{4}{3} \int \frac{\sqrt{3} {sec}^{2} θ d θ}{1 + {tan}^{2} θ}$ $4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/3int(sqrt3 sec^2 theta d theta)/(1+ tan^2theta)$

But $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2 theta=sec^2theta$

$\frac{4}{3} \int \frac{d x}{({(\frac{x + 1}{\sqrt{3}})}^{2} + 1)} = \frac{4}{\sqrt{3}} \int d θ$ $4/3int(dx)/((((x+1)/sqrt3)^2+1))=4/sqrt3intd theta$

$= \frac{4}{\sqrt{3}} arctan (\frac{x + 1}{\sqrt{3}})$ $=4/sqrt3arctan((x+1)/sqrt3)$

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