integral from 3 to 4 of (x+4)/(x^2(x-2))

解

\int_{3}^{4} \frac{x + 4}{x ^{2} ( x - 2 )} d x

- \frac{ln ( 2 ) - ln ( 3 )}{2} - \frac{1}{6} - ln (2) + ln (3)

十進法表記

0.44153 \dots

解答ステップ

\int_{3}^{4} \frac{x + 4}{x ^{2} ( x - 2 )} d x

拡張

\frac{x + 4}{x ^{2} ( x - 2 )} : \frac{x}{x ^{2} ( x - 2 )} + \frac{4}{x ^{2} ( x - 2 )}

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_{3}^{4} \frac{x}{x ^{2} ( x - 2 )} d x + \int_{3}^{4} \frac{4}{x ^{2} ( x - 2 )} d x

\int_{3}^{4} \frac{x}{x ^{2} ( x - 2 )} d x = - \frac{ln ( 2 ) - ln ( 3 )}{2}

\int_{3}^{4} \frac{4}{x ^{2} ( x - 2 )} d x = - \frac{1}{6} - ln (2) + ln (3)

= - \frac{ln ( 2 ) - ln ( 3 )}{2} - \frac{1}{6} - ln (2) + ln (3)