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

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

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integral of 1/(-1-2x)

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

- \frac{1}{2} ln ∣ 1 + 2 x ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = - 1 - 2 x

= - \frac{1}{2} ln ∣ - 1 - 2 x ∣

Apply absolute rule:

∣ - a ∣ = ∣ a ∣

∣ - 1 - 2 x ∣ = ∣ 1 + 2 x ∣

= - \frac{1}{2} ln ∣ 1 + 2 x ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ 1 + 2 x ∣ + C

\frac{d}{d x} ((5 x^{6} + 2 x^{3})^{4})

\frac{1}{( s + 1 ) ^{5}}

n = 0 \sum \infty \frac{1}{2 ^{n + 1}}

\int (3 x^{2} - 2 x)^{3} d x

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

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