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

解

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

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

十進法表記

2.00363 \dots

解答ステップ

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

\frac{x ^{2} + 2}{3 x - x ^{2}} = \frac{3 x + 2}{3 x - x ^{2}} - 1

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

\int_{1}^{2} \frac{3 x + 2}{3 x - x ^{2}} d x = 3 ln (2) + \frac{4 ln ( 2 )}{3}

\int_{1}^{2} 1 d x = 1

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