What is the integral of 3/(8+6x) ?

The integral of 3/(8+6x) is 1/2 ln|8+6x|+C

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integral of 3/(8+6x)

\int \frac{3}{8 + 6 x} d x

\frac{1}{2} ln ∣ 8 + 6 x ∣ + C

Solution steps

\int \frac{3}{8 + 6 x} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{8 + 6 x} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 3 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 8 + 6 x

= 3 \cdot \frac{1}{6} ln ∣ 8 + 6 x ∣

Simplify

3 \cdot \frac{1}{6} ln ∣ 8 + 6 x ∣ : \frac{1}{2} ln ∣ 8 + 6 x ∣

= \frac{1}{2} ln ∣ 8 + 6 x ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 8 + 6 x ∣ + C

y = \frac{1}{x}, y = \frac{1}{4} x

x \to 1 lim (\frac{x + 2}{x ^{3} + 8})

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

\frac{\partial}{\partial x} (x^{4} ln (yz))

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