What is the integral from 1 to e of (12+ln(x))^2 ?

The integral from 1 to e of (12+ln(x))^2 is 145e-122

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integral from 1 to e of (12+ln(x))^2

\int_{1}^{e} (12 + ln (x))^{2} d x

145 e - 122

Decimal Notation

272.15086 \dots

Solution steps

\int_{1}^{e} (12 + ln (x))^{2} d x

12 + ln (x) = 12 (1 + \frac{ln ( x )}{12})

= \int_{1}^{e} (12 (1 + \frac{ln ( x )}{12}))^{2} d x

Simplify

(12 (1 + \frac{ln ( x )}{12}))^{2} : 144 + 24 ln (x) + ln^{2} (x)

= \int_{1}^{e} 144 + 24 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} 144 d x + \int_{1}^{e} 24 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 144 d x = 144 e - 144

\int_{1}^{e} 24 ln (x) d x = 24

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

= 144 e - 144 + 24 + e - 2

Simplify

= 145 e - 122

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What is the integral from 1 to e of (12+ln(x))^2 ?
The integral from 1 to e of (12+ln(x))^2 is 145e-122