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

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

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integral of 1/(3x-3)

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

\frac{1}{3} ln ∣ 3 x - 3 ∣ + C

Solution steps

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

= \frac{1}{3} ln ∣ 3 x - 3 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x - 3 ∣ + C

\int_{4}^{6} \frac{x}{x ^{2} + 6 x + 13} d x

x^{2} 9 - x^{2}

x \to 2 lim (2 x^{3} + 4 e^{x})

- x^{3} + x, (- 1, 1.19148)

\frac{9 u ^{2}}{( u ^{2} + u ) ^{3}}