How do you evaluate the definite integral $\int \frac{x}{x^{2} - 1}$ $int x/(x^2-1)$ from $[2, 3]$ $[2,3]$ ?

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How do you evaluate the definite integral $\int \frac{x}{x^{2} - 1}$ $int x/(x^2-1)$ from $[2, 3]$ $[2,3]$ ?

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Answer 1 · 2016-10-15T07:25:25

1/2ln(8/3)

Explanation:

$\int x \frac{d x}{x^{2} - 1} = \int x \frac{d x}{(x + 1) (x - 1)}$ $intxdx/(x^2-1)=intxdx/((x+1)(x-1))$
$\frac{x}{(x + 1) (x - 1)} = \frac{A}{x + 1} + \frac{B}{x - 1}$ $x/((x+1)(x-1))=A/(x+1)+B/(x-1)$
$x = A (x - 1) + B (x + 1)$ $x=A(x-1)+B(x+1)$
$x = 1 \Rightarrow 1 = 0 + 2 B; B = \frac{1}{2}$ $x=1 =>1=0+2B ; B=1/2$
$x = - 1 \Rightarrow - 1 = 0 - 2 A; A = \frac{1}{2}$ $x=-1 =>-1=0-2A ; A=1/2$
$I = \int x \frac{d x}{x^{2} - 1} = \frac{1}{2} (\int 1 \frac{d x}{x + 1} + \int 1 \frac{d x}{x - 1}) = \frac{1}{2} (ln (x + 1) + ln (x - 1)) = \frac{1}{2} (ln (x + 1) (x - 1)$ $I=intxdx/(x^2-1)=1/2(int1dx/(x+1)+int1dx/(x-1))=1/2(ln(x+1)+ln(x-1))=1/2(ln(x+1)(x-1)$
$x = 3; I = \frac{1}{2} ln 8$ $x=3 ; I=1/2Ln8$
$x = 2; I = \frac{1}{2} ln 3$ $x=2 ; I=1/2Ln3$
$I = \frac{1}{2} (ln 8 - ln 3) = \frac{1}{2} ln (\frac{8}{3})$ $I=1/2(ln8-ln3)=1/2ln(8/3)$

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How do you evaluate the definite integral $\int \frac{x}{x^{2} - 1}$ $int x/(x^2-1)$ from $[2, 3]$ $[2,3]$ ?

1 Answer

Explanation:

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How do you evaluate the definite integral ∫xx2−1int x/(x^2-1) from [2,3][2,3]?

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How do you evaluate the definite integral $\int \frac{x}{x^{2} - 1}$ $int x/(x^2-1)$ from $[2, 3]$ $[2,3]$ ?