integral from 1 to 4 of (x^2+2)/(5x-x^2)

解

\int_{1}^{4} \frac{x ^{2} + 2}{5 x - x ^{2}} d x

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

十進法表記

5.04050 \dots

解答ステップ

\int_{1}^{4} \frac{x ^{2} + 2}{5 x - x ^{2}} d x

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

= \int_{1}^{4} \frac{5 x + 2}{5 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}^{4} \frac{5 x + 2}{5 x - x ^{2}} d x - \int_{1}^{4} 1 d x

\int_{1}^{4} \frac{5 x + 2}{5 x - x ^{2}} d x = 10 ln (2) + \frac{8 ln ( 2 )}{5}

\int_{1}^{4} 1 d x = 3

= 10 ln (2) + \frac{8 ln ( 2 )}{5} - 3