How do you integrate $ln (x^{2} - 1) d x$ $ln(x^2 -1)dx$ ?

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Answer 1 · 2016-07-31T20:38:03

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

we can try going with IBP: $int u v' = uv - int u'v$

so

$int \ ln(x^2 -1) \ dx$

$= int \d/dx(x) ln(x^2 -1) \ dx$

and by IBP

$= xln(x^2 -1) - int \xd/dx (ln(x^2 -1)) \ dx$

$= xln(x^2 -1) - int \x 1/(x^2 -1)* 2x \ dx$

$= xln(x^2 -1) - 2 int \x^2/(x^2 -1) \ dx$

$= xln(x^2 -1) - 2 int \ (x^2- 1 + 1)/(x^2 -1) \ dx$

$= xln(x^2 -1) - 2 int \ 1 + 1/(x^2 -1) \ dx$

$= xln(x^2 -1) - 2x - 2 int \ 1/(x^2 -1) \ dx$

$= xln(x^2 -1) - 2x - 2 int \ 1/((x -1)(x+1)) \ dx$

so now using partial fractions

$1/((x -1)(x+1)) = 1/2( 1/(x-1) - 1/(x+1))$

$implies xln(x^2 -1) - 2x - int \ 1/(x-1) - 1/(x+1) \ dx$

$= xln(x^2 -1) - 2x - ln (x-1) + ln (x+1) + C$

$= xln(x^2 -1) - 2x + ln ((x+1)/(x-1)) + C$

How do you integrate ln(x^2 -1)dxln(x2−1)dx ln(x^2 -1)dx?