What is the integral from 1 to e of (24+ln(x))^2 ?

The integral from 1 to e of (24+ln(x))^2 is 577e-530

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integral from 1 to e of (24+ln(x))^2

\int_{1}^{e} (24 + ln (x))^{2} d x

577 e - 530

Decimal

1038.44861 \dots

Solution steps

\int_{1}^{e} (24 + ln (x))^{2} d x

24 + ln (x) = 24 (1 + \frac{ln ( x )}{24})

= \int_{1}^{e} (24 (1 + \frac{ln ( x )}{24}))^{2} d x

Simplify

(24 (1 + \frac{ln ( x )}{24}))^{2} : 576 + 48 ln (x) + ln^{2} (x)

= \int_{1}^{e} 576 + 48 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} 576 d x + \int_{1}^{e} 48 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 576 d x = 576 e - 576

\int_{1}^{e} 48 ln (x) d x = 48

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

= 576 e - 576 + 48 + e - 2

Simplify

= 577 e - 530

What is the integral from 1 to e of (24+ln(x))^2 ?
The integral from 1 to e of (24+ln(x))^2 is 577e-530