What is the integral from 2 to 3 of (1/(x-1)+x) ?

The integral from 2 to 3 of (1/(x-1)+x) is ln(2)+5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 3 of (1/(x-1)+x)

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

ln (2) + \frac{5}{2}

Decimal

3.19314 \dots

Solution steps

\int_{2}^{3} \frac{1}{x - 1} + x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{3} \frac{1}{x - 1} d x + \int_{2}^{3} x d x

\int_{2}^{3} \frac{1}{x - 1} d x = ln (2)

\int_{2}^{3} x d x = \frac{5}{2}

= ln (2) + \frac{5}{2}

\int_{6}^{5} (sin (x)) d x

\int_{2}^{8} - \frac{6}{x} d x

\int_{1}^{3} 9 x^{2} d x

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

\int_{1}^{2} (0.5 - 0.125 x) d x

What is the integral from 2 to 3 of (1/(x-1)+x) ?
The integral from 2 to 3 of (1/(x-1)+x) is ln(2)+5/2