What is the integral from 1 to 2 of ln((4x)^2) ?

The integral from 1 to 2 of ln((4x)^2) is 8ln(2)-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of ln((4x)^2)

\int_{1}^{2} ln ((4 x)^{2}) d x

8 ln (2) - 2

Decimal Notation

3.54517 \dots

Solution steps

\int_{1}^{2} ln ((4 x)^{2}) d x

Apply Integration By Parts

: [x ln (16 x^{2}) - \int 2 d x]_{1}^{2}

= [x ln (16 x^{2}) - \int 2 d x]_{1}^{2}

\int 2 d x = 2 x

= [x ln (16 x^{2}) - 2 x]_{1}^{2}

Simplify

ln (16 x^{2}) : 2 (ln (x) + 2 ln (2))

= [2 x (ln (x) + 2 ln (2)) - 2 x]_{1}^{2}

Compute the boundaries:

8 ln (2) - 2

= 8 ln (2) - 2

\int_{1}^{6} x^{4} ln (x) d x

\int_{0}^{\frac{y}{4}} e^{- y^{2}} d y

\int_{0}^{1} (3 - 2 x)^{- 2} d x

\int_{0}^{1} x^{2} \cdot e^{- 10 x} d x

\int_{1}^{\infty} \frac{1}{2 x + 9} d x

What is the integral from 1 to 2 of ln((4x)^2) ?
The integral from 1 to 2 of ln((4x)^2) is 8ln(2)-2