integral from 1 to 2 of (4y^2-6y-12)/(y(y+2)(y-3))

解

\int_{1}^{2} \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

2 ln (2) + \frac{14 ln ( 2 ) - 8 ln ( 3 )}{5}

十進法表記

1.56932 \dots

解答ステップ

\int_{1}^{2} \frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} d y

以下の部分分数を得る:

\frac{4 y ^{2} - 6 y - 12}{y ( y + 2 ) ( y - 3 )} : \frac{2}{y} + \frac{8}{5 ( y + 2 )} + \frac{2}{5 ( y - 3 )}

= \int_{1}^{2} \frac{2}{y} + \frac{8}{5 ( y + 2 )} + \frac{2}{5 ( y - 3 )} d y

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}^{2} \frac{2}{y} d y + \int_{1}^{2} \frac{8}{5 ( y + 2 )} d y + \int_{1}^{2} \frac{2}{5 ( y - 3 )} d y

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

\int_{1}^{2} \frac{8}{5 ( y + 2 )} d y = \frac{8}{5} (2 ln (2) - ln (3))

\int_{1}^{2} \frac{2}{5 ( y - 3 )} d y = - \frac{2}{5} ln (2)

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

簡素化

2 ln (2) + \frac{8}{5} (2 ln (2) - ln (3)) - \frac{2}{5} ln (2) : 2 ln (2) + \frac{14 ln ( 2 ) - 8 ln ( 3 )}{5}

= 2 ln (2) + \frac{14 ln ( 2 ) - 8 ln ( 3 )}{5}