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

The integral from 1 to e of (x^2+1)/x is (e^2-1)/2+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 (x^2+1)/x

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

\frac{e ^{2} - 1}{2} + 1

Decimal Notation

4.19452 \dots

Solution steps

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

Expand

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

= \int_{1}^{e} 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}^{e} x d x + \int_{1}^{e} \frac{1}{x} d x

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

\int_{1}^{e} \frac{1}{x} d x = 1

= \frac{e ^{2} - 1}{2} + 1

y^{^{''}} = - 25 y

\int_{0}^{1} (x^{2} - x) d x

\int \frac{1}{6 - x} d x

\int 1 + x^{4} d x

\frac{d}{d x} (ln ∣ sec (x) ∣)

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