integral from 1 to 3 of 1+1/x+1/(x^2)

解

\int_{1}^{3} 1 + \frac{1}{x} + \frac{1}{x ^{2}} d x

ln (3) + \frac{8}{3}

十進法表記

3.76527 \dots

解答ステップ

\int_{1}^{3} 1 + \frac{1}{x} + \frac{1}{x ^{2}} d x

総和規則を適用する:

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

= \int_{1}^{3} 1 d x + \int_{1}^{3} \frac{1}{x} d x + \int_{1}^{3} \frac{1}{x ^{2}} d x

\int_{1}^{3} 1 d x = 2

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

\int_{1}^{3} \frac{1}{x ^{2}} d x = \frac{2}{3}

= 2 + ln (3) + \frac{2}{3}

分数を組み合わせる

2 + \frac{2}{3} : \frac{8}{3}

= ln (3) + \frac{8}{3}