integral from 0 to 3 of 1/((x+5)(x^2+9))

解

\int_{0}^{3} \frac{1}{( x + 5 ) ( x ^{2} + 9 )} d x

\frac{30 ln ( 2 ) + 5 π - 12 ln ( 5 )}{408}

十進法表記

0.04213 \dots

解答ステップ

\int_{0}^{3} \frac{1}{( x + 5 ) ( x ^{2} + 9 )} d x

以下の部分分数を得る:

\frac{1}{( x + 5 ) ( x ^{2} + 9 )} : \frac{1}{34 ( x + 5 )} + \frac{- x + 5}{34 ( x ^{2} + 9 )}

= \int_{0}^{3} \frac{1}{34 ( x + 5 )} + \frac{- x + 5}{34 ( x ^{2} + 9 )} 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

= \int_{0}^{3} \frac{1}{34 ( x + 5 )} d x + \int_{0}^{3} \frac{- x + 5}{34 ( x ^{2} + 9 )} d x

\int_{0}^{3} \frac{1}{34 ( x + 5 )} d x = \frac{1}{34} (3 ln (2) - ln (5))

\int_{0}^{3} \frac{- x + 5}{34 ( x ^{2} + 9 )} d x = \frac{1}{102} (- \frac{3}{2} ln (2) + \frac{5 π}{4})

= \frac{1}{34} (3 ln (2) - ln (5)) + \frac{1}{102} (- \frac{3}{2} ln (2) + \frac{5 π}{4})

簡素化

\frac{1}{34} (3 ln (2) - ln (5)) + \frac{1}{102} (- \frac{3}{2} ln (2) + \frac{5 π}{4}) : \frac{30 ln ( 2 ) + 5 π - 12 ln ( 5 )}{408}

= \frac{30 ln ( 2 ) + 5 π - 12 ln ( 5 )}{408}