What is the integral of (ln(4x))/(4x^2) ?

The integral of (ln(4x))/(4x^2) is (-ln(4x)-1)/(4x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (ln(4x))/(4x^2)

\int \frac{ln ( 4 x )}{4 x ^{2}} d x

\frac{- ln ( 4 x ) - 1}{4 x} + C

Solution steps

\int \frac{ln ( 4 x )}{4 x ^{2}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{ln ( 4 x )}{x ^{2}} d x

Apply Integration By Parts

: - \frac{ln ( 4 x )}{x} - \int - \frac{1}{x ^{2}} d x

= \frac{1}{4} (- \frac{ln ( 4 x )}{x} - \int - \frac{1}{x ^{2}} d x)

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

= \frac{1}{4} (- \frac{ln ( 4 x )}{x} - \frac{1}{x})

Simplify

\frac{1}{4} (- \frac{ln ( 4 x )}{x} - \frac{1}{x}) : \frac{- ln ( 4 x ) - 1}{4 x}

= \frac{- ln ( 4 x ) - 1}{4 x}

Add a constant to the solution

= \frac{- ln ( 4 x ) - 1}{4 x} + C

e^{x^{2} - 9}

2 x^{4} - x^{3} + 6 x^{2}

a (t) = t^{3} - 12 t

\int_{0}^{\infty} x^{3} e^{- 3 x} d x

\int + x^{2} a^{2} - x^{2} d x

What is the integral of (ln(4x))/(4x^2) ?
The integral of (ln(4x))/(4x^2) is (-ln(4x)-1)/(4x)+C