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

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

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

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

arcsin (x + 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 1} d u

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

Use the common integral:

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

= arcsin (u)

Substitute back

u = x + 1

= arcsin (x + 1)

Add a constant to the solution

= arcsin (x + 1) + C

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What is the integral of 1/(sqrt(-x^2-2x)) ?
The integral of 1/(sqrt(-x^2-2x)) is arcsin(x+1)+C