integral from 1 to 3 of (x+3)/(x^2+6x)

解

\int_{1}^{3} \frac{x + 3}{x ^{2} + 6 x} d x

\frac{3 ln ( 3 ) - ln ( 7 )}{2}

十進法表記

0.67496 \dots

解答ステップ

\int_{1}^{3} \frac{x + 3}{x ^{2} + 6 x} d x

以下の部分分数を得る:

\frac{x + 3}{x ^{2} + 6 x} : \frac{1}{2 x} + \frac{1}{2 ( x + 6 )}

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

\int_{1}^{3} \frac{1}{2 x} d x = \frac{1}{2} ln (3)

\int_{1}^{3} \frac{1}{2 ( x + 6 )} d x = ln (3) - \frac{1}{2} ln (7)

= \frac{1}{2} ln (3) + ln (3) - \frac{1}{2} ln (7)

簡素化

\frac{1}{2} ln (3) + ln (3) - \frac{1}{2} ln (7) : \frac{3 ln ( 3 ) - ln ( 7 )}{2}

= \frac{3 ln ( 3 ) - ln ( 7 )}{2}