What is the integral of 1/(x(2+3u)) ?

The integral of 1/(x(2+3u)) is 1/(2+3u)ln|x|+C

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integral of 1/(x(2+3u))

\int \frac{1}{x ( 2 + 3 u )} d x

\frac{1}{2 + 3 u} ln ∣ x ∣ + C

Solution steps

\int \frac{1}{x ( 2 + 3 u )} d x

Take the constant out:

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

= \frac{1}{2 + 3 u} \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{2 + 3 u} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{2 + 3 u} ln ∣ x ∣ + C

y^{^{''}} - 2 y^{^{'}} + y = 0

\int x (12 x + 4) d x

x \to 2 lim (ln ∣ 2 - x ∣)

f (x) = - x^{3} + 9 x^{2} - 6 x + 9, at x = 7

x \to - \infty lim (\frac{x}{c})

What is the integral of 1/(x(2+3u)) ?
The integral of 1/(x(2+3u)) is 1/(2+3u)ln|x|+C