How do you find $\int \frac{x - 1}{(x^{3} + 1)} d x$ $int (x-1)/((x^3+1))dx$ using partial fractions?

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How do you find $\int \frac{x - 1}{(x^{3} + 1)} d x$ $int (x-1)/((x^3+1))dx$ using partial fractions?

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Answer 1 · 2015-11-15T22:05:09

$I = - \frac{2}{3} ln | x + 1 | + \frac{1}{3} ln ∣ ∣ x^{2} - x + 1 ∣ ∣ + C$ $I=-2/3ln|x+1|+1/3ln|x^2-x+1|+C$

Explanation:

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

$\frac{x - 1}{(x + 1) (x^{2} - x + 1)} = \frac{A}{x + 1} + \frac{B x + C}{x^{2} - x + 1} =$ $(x-1)/((x+1)(x^2-x+1))=A/(x+1)+(Bx+C)/(x^2-x+1)=$

$= \frac{A (x^{2} - x + 1) + (B x + C) (x + 1)}{(x + 1) (x^{2} - x + 1)} =$ $=(A(x^2-x+1)+(Bx+C)(x+1))/((x+1)(x^2-x+1))=$

$= \frac{A x^{2} - A x + A + B x^{2} + B x + C x + C}{(x + 1) (x^{2} - x + 1)} =$ $=(Ax^2-Ax+A+Bx^2+Bx+Cx+C)/((x+1)(x^2-x+1))=$

$= \frac{x^{2} (A + B) + x (- A + B + C) + (A + C)}{(x + 1) (x^{2} - x + 1)}$ $=(x^2(A+B)+x(-A+B+C)+(A+C))/((x+1)(x^2-x+1))$

$A + B = 0$ $A+B=0$
$- A + B + C = 1$ $-A+B+C=1$
$A + C = - 1$ $A+C=-1$

$2 B + C = 1$ $2B+C=1$
$B + 2 C = 0$ $B+2C=0$

$- 3 B = - 2 \Rightarrow B = \frac{2}{3}$ $-3B=-2 => B=2/3$

$C = 1 - 2 B = - \frac{1}{3}$ $C=1-2B = -1/3$

$A = - B = - \frac{2}{3}$ $A=-B = -2/3$

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

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

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

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

$I = - \frac{2}{3} \int \frac{d x}{x + 1} + \frac{1}{3} \int \frac{d (x^{2} - x + 1)}{x^{2} - x + 1}$ $I=-2/3intdx/(x+1) + 1/3int (d(x^2-x+1))/(x^2-x+1)$

$I = - \frac{2}{3} ln | x + 1 | + \frac{1}{3} ln ∣ ∣ x^{2} - x + 1 ∣ ∣ + C$ $I=-2/3ln|x+1|+1/3ln|x^2-x+1|+C$

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How do you find $\int \frac{x - 1}{(x^{3} + 1)} d x$ $int (x-1)/((x^3+1))dx$ using partial fractions?

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Explanation:

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How do you find $\int \frac{x - 1}{(x^{3} + 1)} d x$ $int (x-1)/((x^3+1))dx$ using partial fractions?