What is the integral from 1 to e of (11+ln(x))^2 ?

The integral from 1 to e of (11+ln(x))^2 is 122e-101

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integral from 1 to e of (11+ln(x))^2

\int_{1}^{e} (11 + ln (x))^{2} d x

122 e - 101

Decimal Notation

230.63038 \dots

Solution steps

\int_{1}^{e} (11 + ln (x))^{2} d x

Expand

(11 + ln (x))^{2} : 121 + 22 ln (x) + ln^{2} (x)

= \int_{1}^{e} 121 + 22 ln (x) + ln^{2} (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} 121 d x + \int_{1}^{e} 22 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 121 d x = 121 e - 121

\int_{1}^{e} 22 ln (x) d x = 22

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

= 121 e - 121 + 22 + e - 2

Simplify

= 122 e - 101

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What is the integral from 1 to e of (11+ln(x))^2 ?
The integral from 1 to e of (11+ln(x))^2 is 122e-101