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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

Add a constant to the solution

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

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