integral from 1 to 2 of (13)/(x^3+5x)

解

\int_{1}^{2} \frac{13}{x ^{3} + 5 x} d x

13 (\frac{1}{5} ln (2) - \frac{1}{10} ln (\frac{3}{2}))

十進法表記

1.27507 \dots

解答ステップ

\int_{1}^{2} \frac{13}{x ^{3} + 5 x} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 13 \cdot \int_{1}^{2} \frac{1}{x ^{3} + 5 x} d x

以下の部分分数を得る:

\frac{1}{x ^{3} + 5 x} : \frac{1}{5 x} - \frac{x}{5 ( x ^{2} + 5 )}

= 13 \cdot \int_{1}^{2} \frac{1}{5 x} - \frac{x}{5 ( x ^{2} + 5 )} 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

= 13 (\int_{1}^{2} \frac{1}{5 x} d x - \int_{1}^{2} \frac{x}{5 ( x ^{2} + 5 )} d x)

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

\int_{1}^{2} \frac{x}{5 ( x ^{2} + 5 )} d x = \frac{1}{10} ln (\frac{3}{2})

= 13 (\frac{1}{5} ln (2) - \frac{1}{10} ln (\frac{3}{2}))