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

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

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

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

- arctan (x) + C

Solution steps

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

Factor

- x^{2} - 1 : - (x^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

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

= - arctan (x)

Add a constant to the solution

= - arctan (x) + C

y^{^{'}} = \frac{( x y ^{3} )}{25}

\frac{d y}{d x} = \frac{6 y}{x ^{2}}

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\int csc^{2} (x) + 5 d x