integral from 1 to 6 of (x^2+2)/(7x-x^2)

解

\int_{1}^{6} \frac{x ^{2} + 2}{7 x - x ^{2}} d x

7 ln (6) + \frac{4 ln ( 6 )}{7} - 5

十進法表記

8.56617 \dots

解答ステップ

\int_{1}^{6} \frac{x ^{2} + 2}{7 x - x ^{2}} d x

\frac{x ^{2} + 2}{7 x - x ^{2}} = \frac{7 x + 2}{7 x - x ^{2}} - 1

= \int_{1}^{6} \frac{7 x + 2}{7 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}^{6} \frac{7 x + 2}{7 x - x ^{2}} d x - \int_{1}^{6} 1 d x

\int_{1}^{6} \frac{7 x + 2}{7 x - x ^{2}} d x = 7 ln (6) + \frac{4 ln ( 6 )}{7}

\int_{1}^{6} 1 d x = 5

= 7 ln (6) + \frac{4 ln ( 6 )}{7} - 5