What is the integral of (1+x)/(-4x-2x^2) ?

The integral of (1+x)/(-4x-2x^2) is -1/4 ln|4x+2x^2|+C

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integral of (1+x)/(-4x-2x^2)

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

- \frac{1}{4} ln 4 x + 2 x^{2} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = - 4 x - 2 x^{2}

= - \frac{1}{4} ln - 4 x - 2 x^{2}

Apply absolute rule:

∣ - a ∣ = ∣ a ∣

- 4 x - 2 x^{2} = 4 x + 2 x^{2}

= - \frac{1}{4} ln 4 x + 2 x^{2}

Add a constant to the solution

= - \frac{1}{4} ln 4 x + 2 x^{2} + C

\frac{d}{d x} (\frac{e ^{5 x}}{x ^{15}})

\frac{d}{d x} (b (x + a)^{3} + c)

\frac{\partial}{\partial x} (x y e^{\frac{x}{y}})

x \to \infty lim (x \cdot 2)

What is the integral of (1+x)/(-4x-2x^2) ?
The integral of (1+x)/(-4x-2x^2) is -1/4 ln|4x+2x^2|+C