integral from 1 to e of (13+ln(x))^2

解答

\int_{1}^{e} (13 + ln (x))^{2} d x

170 e - 145

十进制

317.10791 \dots

求解步骤

\int_{1}^{e} (13 + ln (x))^{2} d x

13 + ln (x) = 13 (1 + \frac{ln ( x )}{13})

= \int_{1}^{e} (13 (1 + \frac{ln ( x )}{13}))^{2} d x

化简

(13 (1 + \frac{ln ( x )}{13}))^{2} : 169 + 26 ln (x) + ln^{2} (x)

= \int_{1}^{e} 169 + 26 ln (x) + ln^{2} (x) d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{e} 169 d x + \int_{1}^{e} 26 ln (x) d x + \int_{1}^{e} ln^{2} (x) d x

\int_{1}^{e} 169 d x = 169 e - 169

\int_{1}^{e} 26 ln (x) d x = 26

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

= 169 e - 169 + 26 + e - 2

化简

= 170 e - 145

\int_{0}^{\infty} Q V d r

\int_{0}^{2} \frac{2}{6 - x ^{2}} d x

\int_{3}^{6} \frac{4 x - 4}{x ^{2} - 5 x} d x

\int_{0}^{s i n (x)} e^{y} cos (x) d y

\int_{0}^{\frac{1}{12}} x ln (12 x) d x