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

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

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

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

2 arcsin (x + 1) + C

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

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

= 2 arcsin (u)

Substitute back

u = x + 1

= 2 arcsin (x + 1)

Add a constant to the solution

= 2 arcsin (x + 1) + C

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

\int \frac{1}{- x ^{2} - 4 x + 5} d x

\int \frac{2}{x ( ln ( x ) ) ^{4}} d x

\int \frac{x + 1}{x ^{2} + 4 x + 13} d x

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

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