What is the integral of 1/(9-3x) ?

The integral of 1/(9-3x) is -1/3 ln|9-3x|+C

Popular >

integral of 1/(9-3x)

\int \frac{1}{9 - 3 x} d x

- \frac{1}{3} ln ∣ 9 - 3 x ∣ + C

Solution steps

\int \frac{1}{9 - 3 x} 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 = 9 - 3 x

= - \frac{1}{3} ln ∣ 9 - 3 x ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ 9 - 3 x ∣ + C

y^{^{''}} - 2 y^{^{'}} + y = 2 (1 - x) \cdot cos (x) - 2 sin (x)

3 x^{2}, x^{4} - x^{2}, [- 2, 2]

\int \frac{30 x ^{3} - 60 x ^{2} + 7}{x ^{2} - 2 x} d x

\int \frac{- 15}{x ^{2} x ^{2} + 4} d x

\int 3 \frac{x ^{4}}{e ^{x^{5}}} d x