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)

\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

y^{^{'}} = 2 y + x^{2} + 7

\frac{d ^{2}}{d x ^{2}} (e^{3 x})

\frac{1}{s ^{2} + 8} (- \frac{40}{33})

\frac{d}{d x} (- 0.025 x^{2} + 9 x)

\frac{7}{2 7 x + 2}, at a = 2

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