What is the integral from 1 to e of (23+ln(x))^2 ?

The integral from 1 to e of (23+ln(x))^2 is 530e-485

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integral from 1 to e of (23+ln(x))^2

\int_{1}^{e} (23 + ln (x))^{2} d x

530 e - 485

Decimal Notation

955.68936 \dots

Solution steps

\int_{1}^{e} (23 + ln (x))^{2} d x

23 + ln (x) = 23 (1 + \frac{ln ( x )}{23})

= \int_{1}^{e} (23 (1 + \frac{ln ( x )}{23}))^{2} d x

Simplify

(23 (1 + \frac{ln ( x )}{23}))^{2} : 529 + 46 ln (x) + ln^{2} (x)

= \int_{1}^{e} 529 + 46 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} 529 d x + \int_{1}^{e} 46 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 529 d x = 529 e - 529

\int_{1}^{e} 46 ln (x) d x = 46

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

= 529 e - 529 + 46 + e - 2

Simplify

= 530 e - 485

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What is the integral from 1 to e of (23+ln(x))^2 ?
The integral from 1 to e of (23+ln(x))^2 is 530e-485