What is the integral from 1 to e of (21+ln(x))^2 ?

The integral from 1 to e of (21+ln(x))^2 is 442e-401

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integral from 1 to e of (21+ln(x))^2

\int_{1}^{e} (21 + ln (x))^{2} d x

442 e - 401

Decimal

800.48056 \dots

Solution steps

\int_{1}^{e} (21 + ln (x))^{2} d x

21 + ln (x) = 21 (1 + \frac{ln ( x )}{21})

= \int_{1}^{e} (21 (1 + \frac{ln ( x )}{21}))^{2} d x

Simplify

(21 (1 + \frac{ln ( x )}{21}))^{2} : 441 + 42 ln (x) + ln^{2} (x)

= \int_{1}^{e} 441 + 42 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} 441 d x + \int_{1}^{e} 42 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 441 d x = 441 e - 441

\int_{1}^{e} 42 ln (x) d x = 42

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

= 441 e - 441 + 42 + e - 2

Simplify

= 442 e - 401

What is the integral from 1 to e of (21+ln(x))^2 ?
The integral from 1 to e of (21+ln(x))^2 is 442e-401