What is the integral of (5x+6)^{-1} ?

The integral of (5x+6)^{-1} is 1/5 ln|5x+6|+C

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integral of (5x+6)^{-1}

\int (5 x + 6)^{- 1} d x

\frac{1}{5} ln ∣ 5 x + 6 ∣ + C

Solution steps

\int (5 x + 6)^{- 1} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \int \frac{1}{5 x + 6} d x

Apply u-substitution

: \int \frac{1}{5 u} d u

= \int \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} ln ∣ u ∣

Substitute back

u = 5 x + 6

= \frac{1}{5} ln ∣ 5 x + 6 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ 5 x + 6 ∣ + C

(3 x - 1) d x + (5 y + 7) d y = 0

(6, - 12), (15, - 3)

x \to 0.001 lim (\frac{e ^{x} - 1}{x})

x y \frac{d y}{d x} = ln (x)

\int \frac{x ^{2}}{( 1 - x ^{3} ) ^{2}} d x

What is the integral of (5x+6)^{-1} ?
The integral of (5x+6)^{-1} is 1/5 ln|5x+6|+C