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

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

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integral of 1/(4x+2)

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

\frac{1}{4} ln ∣ 4 x + 2 ∣ + C

Solution steps

\int \frac{1}{4 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

= \frac{1}{4} ln ∣ 4 x + 2 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ 4 x + 2 ∣ + C

k = 0 \sum \infty (2 k + 1) x^{2 k}

y^{^{'}} - y^{2} = 1

n = 1 \sum \infty (- \frac{1}{4})^{n}

\frac{d y}{d x} - \frac{y}{x} = 8 x e^{x}

y = 3 sin (x), y = e^{x}, x = 0, x = \frac{π}{2}