What is the integral of (ln(-2x))/(5x) ?

The integral of (ln(-2x))/(5x) is 1/5 (ln(-2x)ln|x|-(ln|x|^2)/2)+C

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integral of (ln(-2x))/(5x)

\int \frac{ln ( - 2 x )}{5 x} d x

\frac{1}{5} (ln (- 2 x) ln ∣ x ∣ - \frac{ln ∣ x ∣ ^{2}}{2}) + C

Solution steps

\int \frac{ln ( - 2 x )}{5 x} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{ln ( - 2 x )}{x} d x

Apply Integration By Parts

: ln (- 2 x) ln ∣ x ∣ - \int \frac{ln ∣ x ∣}{x} d x

= \frac{1}{5} (ln (- 2 x) ln ∣ x ∣ - \int \frac{ln ∣ x ∣}{x} d x)

\int \frac{ln ∣ x ∣}{x} d x = \frac{ln ∣ x ∣ ^{2}}{2}

= \frac{1}{5} (ln (- 2 x) ln ∣ x ∣ - \frac{ln ∣ x ∣ ^{2}}{2})

Add a constant to the solution

= \frac{1}{5} (ln (- 2 x) ln ∣ x ∣ - \frac{ln ∣ x ∣ ^{2}}{2}) + C

t \to 0 lim (\frac{e ^{t} - 1}{t ^{9}})

y = x^{2} - 7, y = 6 x

6 y^{^{'''}} + 4 y^{^{''}} = 0

\int \frac{x}{e ^{x^{2}}} d x

\int \frac{1 + x ^{2}}{1 + x ^{4}} d x

What is the integral of (ln(-2x))/(5x) ?
The integral of (ln(-2x))/(5x) is 1/5 (ln(-2x)ln|x|-(ln|x|^2)/2)+C