integral from 2 to e+1 of x/(x+1)

解答

\int_{2}^{e + 1} \frac{x}{x + 1} d x

e - 1 - ln (e + 2) + ln (3)

十进制

1.26544 \dots

求解步骤

\int_{2}^{e + 1} \frac{x}{x + 1} d x

使用换元积分法

= \int_{3}^{e + 2} \frac{u - 1}{u} d u

乘开

\frac{u - 1}{u} : 1 - \frac{1}{u}

= \int_{3}^{e + 2} 1 - \frac{1}{u} d u

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_{3}^{e + 2} 1 d u - \int_{3}^{e + 2} \frac{1}{u} d u

\int_{3}^{e + 2} 1 d u = e - 1

\int_{3}^{e + 2} \frac{1}{u} d u = ln (e + 2) - ln (3)

= e - 1 - (ln (e + 2) - ln (3))

化简

= e - 1 - ln (e + 2) + ln (3)

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