What is the integral from 1 to e of (16+ln(x))^2 ?

The integral from 1 to e of (16+ln(x))^2 is 257e-226

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integral from 1 to e of (16+ln(x))^2

\int_{1}^{e} (16 + ln (x))^{2} d x

257 e - 226

Decimal Notation

472.59842 \dots

Solution steps

\int_{1}^{e} (16 + ln (x))^{2} d x

16 + ln (x) = 16 (1 + \frac{ln ( x )}{16})

= \int_{1}^{e} (16 (1 + \frac{ln ( x )}{16}))^{2} d x

Simplify

(16 (1 + \frac{ln ( x )}{16}))^{2} : 256 + 32 ln (x) + ln^{2} (x)

= \int_{1}^{e} 256 + 32 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} 256 d x + \int_{1}^{e} 32 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 256 d x = 256 e - 256

\int_{1}^{e} 32 ln (x) d x = 32

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

= 256 e - 256 + 32 + e - 2

Simplify

= 257 e - 226

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What is the integral from 1 to e of (16+ln(x))^2 ?
The integral from 1 to e of (16+ln(x))^2 is 257e-226