integral from 1 to 5 of (x-1)/(x^2(x+1))

解

\int_{1}^{5} \frac{x - 1}{x ^{2} ( x + 1 )} d x

2 ln (5) - \frac{4}{5} - 2 ln (3)

十進法表記

0.22165 \dots

解答ステップ

\int_{1}^{5} \frac{x - 1}{x ^{2} ( x + 1 )} d x

以下の部分分数を得る:

\frac{x - 1}{x ^{2} ( x + 1 )} : \frac{2}{x} - \frac{1}{x ^{2}} - \frac{2}{x + 1}

= \int_{1}^{5} \frac{2}{x} - \frac{1}{x ^{2}} - \frac{2}{x + 1} 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}^{5} \frac{2}{x} d x - \int_{1}^{5} \frac{1}{x ^{2}} d x - \int_{1}^{5} \frac{2}{x + 1} d x

\int_{1}^{5} \frac{2}{x} d x = 2 ln (5)

\int_{1}^{5} \frac{1}{x ^{2}} d x = \frac{4}{5}

\int_{1}^{5} \frac{2}{x + 1} d x = 2 ln (3)

= 2 ln (5) - \frac{4}{5} - 2 ln (3)