What is the integral from 1 to e of ln(17x) ?

The integral from 1 to e of ln(17x) is eln(17e)-e-ln(17)+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to e of ln(17x)

\int_{1}^{e} ln (17 x) d x

e ln (17 e) - e - ln (17) + 1

Decimal Notation

5.86825 \dots

Solution steps

\int_{1}^{e} ln (17 x) d x

Apply Integration By Parts

: [x ln (17 x) - \int 1 d x]_{1}^{e}

= [x ln (17 x) - \int 1 d x]_{1}^{e}

\int 1 d x = x

= [x ln (17 x) - x]_{1}^{e}

Compute the boundaries:

e ln (17 e) - e - ln (17) + 1

= e ln (17 e) - e - ln (17) + 1

\frac{d y}{d t} = 5

\frac{d}{d x} (arccos (e^{4 x}))

n = 1 \sum \infty sin (n)

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 4

12 x sin (x)

What is the integral from 1 to e of ln(17x) ?
The integral from 1 to e of ln(17x) is eln(17e)-e-ln(17)+1