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

We have to integrate \(\int \frac{x-1}{x+1}\)

Solution

\(\int \frac{x-1}{x+1}\)

We will split the integrand function

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

Using the linearity of the integral

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

These are standard integrals that we can solve directly

\(\int \frac{x-1}{x+1} = x – 2\ln \left | x+1 \right | + C\)

Answer

\(\int \frac{x-1}{x+1} = x – 2\ln \left | x+1 \right | + C\)

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