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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\int \frac{( ln ^{4} ( x ) )}{x} d x

\int \frac{ln ( x + 1 )}{x} d x

\int \frac{3 x}{( 1 + 9 x ^{2} ) ^{4}} d x

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