How do you evaluate the integral $int (x-1)/(x+1)dx$ ?

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Answer 1 · 2017-02-06T13:28:47

$int (x-1) / (x+1) dx = x -2lnabs(x+1) +C$

Split the integrand function:

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

Using the linearity of the integral:

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

These are standard integrals that we can solve directly:

$int (x-1) / (x+1) dx = x -2lnabs(x+1) +C$

How do you evaluate the integral int (x-1)/(x+1)dxint (x-1)/(x+1)dx?