What is the integral from 1 to e of (19+ln(x))^2 ?

The integral from 1 to e of (19+ln(x))^2 is 362e-325

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integral from 1 to e of (19+ln(x))^2

\int_{1}^{e} (19 + ln (x))^{2} d x

362 e - 325

Decimal

659.01802 \dots

Solution steps

\int_{1}^{e} (19 + ln (x))^{2} d x

Expand

(19 + ln (x))^{2} : 361 + 38 ln (x) + ln^{2} (x)

= \int_{1}^{e} 361 + 38 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} 361 d x + \int_{1}^{e} 38 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 361 d x = 361 e - 361

\int_{1}^{e} 38 ln (x) d x = 38

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

= 361 e - 361 + 38 + e - 2

Simplify

= 362 e - 325

What is the integral from 1 to e of (19+ln(x))^2 ?
The integral from 1 to e of (19+ln(x))^2 is 362e-325