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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1 + tan ( x )}{cos ^{2} ( x )} d x

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

\int x - (- d) x d x

\int \frac{2}{8 - x ^{2}} d x

\int \frac{x}{( x - 1 ) ( 2 - x )} d x

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