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

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

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

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

\frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣ + 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 = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

Substitute back

u = x + 1

= \frac{ln ∣ x + 1 + 1 ∣}{2} - \frac{ln ∣ x + 1 - 1 ∣}{2}

Simplify

\frac{ln ∣ x + 1 + 1 ∣}{2} - \frac{ln ∣ x + 1 - 1 ∣}{2} : \frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣

= \frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣ + C

\int \frac{1}{4 x lo g _{3} ( x )} d x

\frac{\partial}{\partial x} (x tan (y))

3 t cos (t)

\frac{d x}{d t} = x + t

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2})

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