What is the integral of 1/(3x+4) ?

The integral of 1/(3x+4) is 1/3 ln|3x+4|+C

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integral of 1/(3x+4)

\int \frac{1}{3 x + 4} d x

\frac{1}{3} ln ∣ 3 x + 4 ∣ + C

Solution steps

\int \frac{1}{3 x + 4} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 4

= \frac{1}{3} ln ∣ 3 x + 4 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x + 4 ∣ + C

x \to 0 lim (\frac{2}{x ^{3}})

x^{2} \frac{d y}{d x} + y^{2} - x y = 0

arcsec (9 x)

\int \frac{6}{x ln ^{3} ( x )} d x

\frac{\partial}{\partial y} (ln (x^{5} + y^{5}))