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

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

integral

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

\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

\frac{1}{( x ^{2} - 4 )}

\frac{4 x ^{4}}{( 1 + x ^{2} )}

\frac{4 t}{3 t ^{2} - 7}

u^{\frac{1}{2}}

\frac{1}{( t ^{2} + t + 1 )}

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