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

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

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+x+1))/x

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

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

Decimal Notation

5.91280 \dots

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

y^{^{''}} + 4 y^{^{'}} - 12 y = 0

(x^{2} - 7) x

x \to 0 lim (\frac{x}{1})

f (x) = 4 x^{3} - 4 x, (1, 0)

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