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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (ln(2x))/(2x^2)

\int \frac{ln ( 2 x )}{2 x ^{2}} d x

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

Solution steps

\int \frac{ln ( 2 x )}{2 x ^{2}} d x

Take the constant out:

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

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

Apply Integration By Parts

: - \frac{ln ( 2 x )}{x} - \int - \frac{1}{x ^{2}} d x

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

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

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

Simplify

\frac{1}{2} (- \frac{ln ( 2 x )}{x} - \frac{1}{x}) : \frac{- ln ( 2 x ) - 1}{2 x}

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

Add a constant to the solution

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

\int t (arctan (t)) d t

\frac{d}{d p} (3 p (1 - 0.5 p))

\frac{d}{d x} (\frac{- 6}{( x + 1 ) ^{4}})

\frac{\partial}{\partial x} (1 + \frac{c}{x} - \frac{b}{x ^{3}})

k = 1 \sum \infty ln (1 + \frac{1}{k})

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