What is the integral from 1 to e^x of 3ln(x) ?

The integral from 1 to e^x of 3ln(x) is 3(e^xx-e^x+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^x of 3ln(x)

\int_{1}^{e^{x}} 3 ln (x) d x

3 (e^{x} x - e^{x} + 1)

Solution steps

\int_{1}^{e^{x}} 3 ln (x) d x

Take the constant out:

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

= 3 \cdot \int_{1}^{e^{x}} ln (x) d x

Apply Integration By Parts

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

= 3 [x ln (x) - \int 1 d x]_{1}^{e^{x}}

\int 1 d x = x

= 3 [x ln (x) - x]_{1}^{e^{x}}

Compute the boundaries:

e^{x} x - e^{x} + 1

= 3 (e^{x} x - e^{x} + 1)

\int_{1}^{4} e^{- x} d x

\int_{- 1}^{2} - x^{2} d x

\int_{- 2}^{2} - x^{2} + 5 d x

\int_{0}^{1} \frac{1}{2} e^{x} + \frac{3}{4} 2^{x} d x

\int_{- 3}^{7} (∣ x ∣) d x

What is the integral from 1 to e^x of 3ln(x) ?
The integral from 1 to e^x of 3ln(x) is 3(e^xx-e^x+1)