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

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

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integral of 1/(4+2x)

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

\frac{1}{2} ln ∣ 4 + 2 x ∣ + C

Solution steps

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

= \frac{1}{2} ln ∣ 4 + 2 x ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 4 + 2 x ∣ + C

y = x^{3} + 7 x - 5, (1, 3)

\int \frac{8 x ^{3}}{( x ^{2} + 2 ) ^{2}} d x

\int (\frac{8}{x}) d x

\frac{d y}{d x} = \frac{1}{x ^{8}}

y = (3 - 2 x + 4 x^{3}), (- 1, 1)