What is the integral from 1 to e of (22+ln(x))^2 ?

The integral from 1 to e of (22+ln(x))^2 is 485e-442

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integral from 1 to e of (22+ln(x))^2

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

485 e - 442

Decimal Notation

876.36668 \dots

Solution steps

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

Expand

(22 + ln (x))^{2} : 484 + 44 ln (x) + ln^{2} (x)

= \int_{1}^{e} 484 + 44 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} 484 d x + \int_{1}^{e} 44 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 484 d x = 484 e - 484

\int_{1}^{e} 44 ln (x) d x = 44

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

= 484 e - 484 + 44 + e - 2

Simplify

= 485 e - 442

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What is the integral from 1 to e of (22+ln(x))^2 ?
The integral from 1 to e of (22+ln(x))^2 is 485e-442