What is the integral from 1 to e of (15+ln(x))^2 ?

The integral from 1 to e of (15+ln(x))^2 is 226e-197

Popular >

integral from 1 to e of (15+ln(x))^2

\int_{1}^{e} (15 + ln (x))^{2} d x

226 e - 197

Decimal

417.33169 \dots

Solution steps

\int_{1}^{e} (15 + ln (x))^{2} d x

15 + ln (x) = 15 (1 + \frac{ln ( x )}{15})

= \int_{1}^{e} (15 (1 + \frac{ln ( x )}{15}))^{2} d x

Simplify

(15 (1 + \frac{ln ( x )}{15}))^{2} : 225 + 30 ln (x) + ln^{2} (x)

= \int_{1}^{e} 225 + 30 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} 225 d x + \int_{1}^{e} 30 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 225 d x = 225 e - 225

\int_{1}^{e} 30 ln (x) d x = 30

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

= 225 e - 225 + 30 + e - 2

Simplify

= 226 e - 197

\int_{1}^{3} \frac{ln ( x )}{x ^{9}} d x

\int_{1}^{2} a (a + 2)^{2} d a

\int_{0}^{2} x^{2} + x + 2 d x

\int_{0}^{k} x x^{2} + 4 d x

\int_{0}^{\infty} \frac{2}{9 + x ^{2}} d x

What is the integral from 1 to e of (15+ln(x))^2 ?
The integral from 1 to e of (15+ln(x))^2 is 226e-197