What is the integral of (x-1)/(1+x^2) ?

The integral of (x-1)/(1+x^2) is 1/2 ln|1+x^2|-arctan(x)+C

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integral of (x-1)/(1+x^2)

\int \frac{x - 1}{1 + x ^{2}} d x

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

Solution steps

\int \frac{x - 1}{1 + x ^{2}} d x

Expand

\frac{x - 1}{1 + x ^{2}} : \frac{x}{1 + x ^{2}} - \frac{1}{1 + x ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{1 + x ^{2}} d x - \int \frac{1}{1 + x ^{2}} d x

\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln 1 + x^{2}

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

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

Add a constant to the solution

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

What is the integral of (x-1)/(1+x^2) ?
The integral of (x-1)/(1+x^2) is 1/2 ln|1+x^2|-arctan(x)+C