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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of (x^2+1)/x

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

\frac{3}{2} + ln (2)

Decimal

2.19314 \dots

Solution steps

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

Expand

\frac{x ^{2} + 1}{x} : x + \frac{1}{x}

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

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

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

= \frac{3}{2} + ln (2)

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