Question-137206

Question Number 137206 by JulioCesar last updated on 31/Mar/21

Answered by bemath last updated on 31/Mar/21

by parts {\begin{cases} u = \ln (\frac{x - 1}{x + 1}) du = \frac{2}{(x - 1) (x + 1)} dx \\ v = x \end{cases}

I = x \ln (\frac{x - 1}{x + 1}) - \int \frac{2 x}{(x - 1) (x + 1)} dx

I = x \ln (\frac{x - 1}{x + 1}) - [\int \frac{1}{x - 1} dx + \int \frac{1}{x + 1} dx]

I = x \ln (\frac{x - 1}{x + 1}) - \ln (x^{2} - 1) + C

I = (x - 1) \ln (x - 1) + (1 - x) \ln (x + 1) + C

Answered by Ñï= last updated on 31/Mar/21

\int l n \frac{x - 1}{x + 1} d x = \int l n (x - 1) d x - \int l n (x + 1) d x

= (x - 1) l n (x - 1) - (x - 1) - (x + 1) l n (x + 1) + (x + 1) + C

= (x - 1) l n (x - 1) - (x + 1) l n (x + 1) + C

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