What is the integral from 1 to e of (20+ln(x))^2 ?

The integral from 1 to e of (20+ln(x))^2 is 401e-362

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integral from 1 to e of (20+ln(x))^2

\int_{1}^{e} (20 + ln (x))^{2} d x

401 e - 362

Decimal Notation

728.03101 \dots

Solution steps

\int_{1}^{e} (20 + ln (x))^{2} d x

Expand

(20 + ln (x))^{2} : 400 + 40 ln (x) + ln^{2} (x)

= \int_{1}^{e} 400 + 40 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} 400 d x + \int_{1}^{e} 40 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 400 d x = 400 e - 400

\int_{1}^{e} 40 ln (x) d x = 40

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

= 400 e - 400 + 40 + e - 2

Simplify

= 401 e - 362

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What is the integral from 1 to e of (20+ln(x))^2 ?
The integral from 1 to e of (20+ln(x))^2 is 401e-362