What is the integral from 1 to e^x of 6ln(t) ?

The integral from 1 to e^x of 6ln(t) is 6(xe^x-e^x+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to e^x of 6ln(t)

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

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

Solution steps

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

Take the constant out:

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

= 6 \cdot \int_{1}^{e^{x}} ln (t) d t

Apply Integration By Parts

= 6 [t ln (t) - \int 1 d t]_{1}^{e^{x}}

\int 1 d t = t

= 6 [t ln (t) - t]_{1}^{e^{x}}

Compute the boundaries:

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

= 6 (x e^{x} - e^{x} + 1)

What is the integral from 1 to e^x of 6ln(t) ?
The integral from 1 to e^x of 6ln(t) is 6(xe^x-e^x+1)