What is the integral from 1 to e of (18+ln(x))^2 ?

The integral from 1 to e of (18+ln(x))^2 is 325e-290

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integral from 1 to e of (18+ln(x))^2

\int_{1}^{e} (18 + ln (x))^{2} d x

325 e - 290

Decimal Notation

593.44159 \dots

Solution steps

\int_{1}^{e} (18 + ln (x))^{2} d x

18 + ln (x) = 18 (1 + \frac{ln ( x )}{18})

= \int_{1}^{e} (18 (1 + \frac{ln ( x )}{18}))^{2} d x

Simplify

(18 (1 + \frac{ln ( x )}{18}))^{2} : 324 + 36 ln (x) + ln^{2} (x)

= \int_{1}^{e} 324 + 36 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} 324 d x + \int_{1}^{e} 36 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 324 d x = 324 e - 324

\int_{1}^{e} 36 ln (x) d x = 36

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

= 324 e - 324 + 36 + e - 2

Simplify

= 325 e - 290

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What is the integral from 1 to e of (18+ln(x))^2 ?
The integral from 1 to e of (18+ln(x))^2 is 325e-290