What is the integral of (9x+36)^{-1} ?

The integral of (9x+36)^{-1} is 1/9 ln|9x+36|+C

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integral of (9x+36)^{-1}

integral

(9 x + 36)^{- 1}

\frac{1}{9} ln ∣ 9 x + 36 ∣ + C

Solution steps

\int (9 x + 36)^{- 1} d x

Apply exponent rule:

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

= \int \frac{1}{9 x + 36} d x

Apply u-substitution

: \int \frac{1}{9 u} d u

= \int \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = 9 x + 36

= \frac{1}{9} ln ∣ 9 x + 36 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ 9 x + 36 ∣ + C

\frac{y ^{3}}{x ^{4}}

\frac{1}{a ^{2} + x ^{2}}

\frac{1}{x ( 1 - x )}

e^{1 + 2 x}

\frac{1}{( ( t + 1 ) \cdot t )}

What is the integral of (9x+36)^{-1} ?
The integral of (9x+36)^{-1} is 1/9 ln|9x+36|+C