What is the integral from 1 to e of (17+ln(x))^2 ?

The integral from 1 to e of (17+ln(x))^2 is 290e-257

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integral from 1 to e of (17+ln(x))^2

\int_{1}^{e} (17 + ln (x))^{2} d x

290 e - 257

Decimal

531.30173 \dots

Solution steps

\int_{1}^{e} (17 + ln (x))^{2} d x

17 + ln (x) = 17 (1 + \frac{ln ( x )}{17})

= \int_{1}^{e} (17 (1 + \frac{ln ( x )}{17}))^{2} d x

Simplify

(17 (1 + \frac{ln ( x )}{17}))^{2} : 289 + 34 ln (x) + ln^{2} (x)

= \int_{1}^{e} 289 + 34 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} 289 d x + \int_{1}^{e} 34 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 289 d x = 289 e - 289

\int_{1}^{e} 34 ln (x) d x = 34

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

= 289 e - 289 + 34 + e - 2

Simplify

= 290 e - 257

What is the integral from 1 to e of (17+ln(x))^2 ?
The integral from 1 to e of (17+ln(x))^2 is 290e-257