integral from 7 to 12 of y/(y^2-3y-4)

解

\int_{7}^{12} \frac{y}{y ^{2} - 3 y - 4} d y

\frac{ln ( 13 ) + 9 ln ( 2 ) - 4 ln ( 3 )}{5}

十進法表記

0.88176 \dots

解答ステップ

\int_{7}^{12} \frac{y}{y ^{2} - 3 y - 4} d y

以下の部分分数を得る:

\frac{y}{y ^{2} - 3 y - 4} : \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )}

= \int_{7}^{12} \frac{1}{5 ( y + 1 )} + \frac{4}{5 ( y - 4 )} 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_{7}^{12} \frac{1}{5 ( y + 1 )} d y + \int_{7}^{12} \frac{4}{5 ( y - 4 )} d y

\int_{7}^{12} \frac{1}{5 ( y + 1 )} d y = \frac{1}{5} (ln (13) - 3 ln (2))

\int_{7}^{12} \frac{4}{5 ( y - 4 )} d y = \frac{4}{5} (3 ln (2) - ln (3))

= \frac{1}{5} (ln (13) - 3 ln (2)) + \frac{4}{5} (3 ln (2) - ln (3))

簡素化

\frac{1}{5} (ln (13) - 3 ln (2)) + \frac{4}{5} (3 ln (2) - ln (3)) : \frac{ln ( 13 ) + 9 ln ( 2 ) - 4 ln ( 3 )}{5}

= \frac{ln ( 13 ) + 9 ln ( 2 ) - 4 ln ( 3 )}{5}