How do you integrate $\int \frac{2 x - 1}{{(x - 1)}^{3} (x - 2)}$ $int (2x - 1) / [(x - 1)^3 (x - 2)]$ using partial fractions?

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How do you integrate $\int \frac{2 x - 1}{{(x - 1)}^{3} (x - 2)}$ $int (2x - 1) / [(x - 1)^3 (x - 2)]$ using partial fractions?

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Answer 1 · 2017-07-09T16:46:22

I decomposed integrand into basic fractions,

$\frac{2 x - 1}{{(x - 1)}^{3} \cdot (x - 2)} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} + \frac{D}{x - 2}$ $(2x-1)/((x-1)^3*(x-2))=A/(x-1)+B/(x-1)^2+C/(x-1)^3+D/(x-2)$

$(2 x - 1) = A \cdot (x^{3} - 4 x^{2} + 5 x - 2) + B \cdot (x^{2} - 3 x + 2) + C \cdot (x - 2) + D \cdot (x^{3} - 3 x^{2} + 3 x - 1)$ $(2x-1)=A*(x^3-4x^2+5x-2)+B*(x^2-3x+2)+C*(x-2)+D*(x^3-3x^2+3x-1)$

Let $x = 1$ $x=1$ , $- C = 1$ $-C=1$ or $C = - 1$ $C=-1$ .

Let $x = 2$ $x=2$ , $D = 3$ $D=3$ .

Differentiate both sides,

$2 = A \cdot (3 x^{2} - 8 x + 5) + B \cdot (2 x - 3) + C + D \cdot (3 x^{2} - 6 x + 3)$ $2=A*(3x^2-8x+5)+B*(2x-3)+C+D*(3x^2-6x+3)$

Let $x = 1$ $x=1$ , $- B + C = 2$ $-B+C=2$ or $- B - 1 = 2$ $-B-1=2$ . Hence $B = - 3$ $B=-3$

Differentiate both sides,

$0, = A \cdot (6 x - 8) + 2 B + D \cdot (6 x - 6)$ $0,=A*(6x-8)+2B+D*(6x-6)$

Let $x = 1$ $x=1$ , $- 2 A + 2 B = 0$ $-2A+2B=0$ or $A = B = - 3$ $A=B=-3$ .

Thus,

$\int \frac{2 x - 1}{{(x - 1)}^{3} \cdot (x - 2)} d x$ $int(2x-1)/((x-1)^3*(x-2))dx$

$= \int - \frac{3}{x - 1} d x + \int - \frac{3}{{(x - 1)}^{2}} d x + + \int - \frac{1}{{(x - 1)}^{3}} d x + \int \frac{3}{x - 2} d x$ $=int-3/(x-1)dx+int-3/(x-1)^2dx++int-1/(x-1)^3dx+int3/(x-2)dx$

$= - 3 ln (x - 1) + \frac{3}{x - 1} + \frac{1}{2} \cdot {(x - 1)}^{- 2} + 3 ln (x - 2) + C$ $=-3Ln(x-1)+3/(x-1)+1/2*(x-1)^(-2)+3Ln(x-2)+C$

$= \frac{3}{x - 1} + \frac{1}{2} \cdot {(x - 1)}^{- 2} + 3 ln (\frac{x - 2}{x - 1}) + C$ $=3/(x-1)+1/2*(x-1)^(-2)+3Ln((x-2)/(x-1))+C$

Explanation:

I decomposed integrand into basic fractions.

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How do you integrate $\int \frac{2 x - 1}{{(x - 1)}^{3} (x - 2)}$ $int (2x - 1) / [(x - 1)^3 (x - 2)]$ using partial fractions?

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