What is the integral from 1 to e of (25+ln(x))^2 ?

The integral from 1 to e of (25+ln(x))^2 is 626e-577

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integral from 1 to e of (25+ln(x))^2

\int_{1}^{e} (25 + ln (x))^{2} d x

626 e - 577

Decimal

1124.64442 \dots

Solution steps

\int_{1}^{e} (25 + ln (x))^{2} d x

Expand

(25 + ln (x))^{2} : 625 + 50 ln (x) + ln^{2} (x)

= \int_{1}^{e} 625 + 50 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} 625 d x + \int_{1}^{e} 50 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 625 d x = 625 e - 625

\int_{1}^{e} 50 ln (x) d x = 50

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

= 625 e - 625 + 50 + e - 2

Simplify

= 626 e - 577

What is the integral from 1 to e of (25+ln(x))^2 ?
The integral from 1 to e of (25+ln(x))^2 is 626e-577