integral from 0 to 1 of (x+5)/(x^2+5x+6)

解

\int_{0}^{1} \frac{x + 5}{x ^{2} + 5 x + 6} d x

\frac{ln ( 2 ) + 5 ln ( \frac{8}{9} )}{2} + 5 (- 3 ln (2) + 2 ln (3))

十進法表記

0.64103 \dots

解答ステップ

\int_{0}^{1} \frac{x + 5}{x ^{2} + 5 x + 6} d x

拡張

\frac{x + 5}{x ^{2} + 5 x + 6} : \frac{x}{x ^{2} + 5 x + 6} + \frac{5}{x ^{2} + 5 x + 6}

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_{0}^{1} \frac{x}{x ^{2} + 5 x + 6} d x + \int_{0}^{1} \frac{5}{x ^{2} + 5 x + 6} d x

\int_{0}^{1} \frac{x}{x ^{2} + 5 x + 6} d x = \frac{ln ( 2 ) + 5 ln ( \frac{8}{9} )}{2}

\int_{0}^{1} \frac{5}{x ^{2} + 5 x + 6} d x = 5 (- 3 ln (2) + 2 ln (3))

= \frac{ln ( 2 ) + 5 ln ( \frac{8}{9} )}{2} + 5 (- 3 ln (2) + 2 ln (3))