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

解

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

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

十進法表記

6.72958 \dots

解答ステップ

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

\frac{x ^{2} + 2}{6 x - x ^{2}} = \frac{6 x + 2}{6 x - x ^{2}} - 1

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

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

\int_{1}^{5} 1 d x = 4

= 6 ln (5) + \frac{2 ln ( 5 )}{3} - 4