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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \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

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

Add a constant to the solution

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

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