integral from 4 to 6 of x/(x-3)

解

\int_{4}^{6} \frac{x}{x - 3} d x

2 + 3 ln (3)

十進法表記

5.29583 \dots

解答ステップ

\int_{4}^{6} \frac{x}{x - 3} d x

u置換積分法を適用する

= \int_{1}^{3} \frac{u + 3}{u} d u

拡張

\frac{u + 3}{u} : 1 + \frac{3}{u}

= \int_{1}^{3} 1 + \frac{3}{u} d u

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} 1 d u + \int_{1}^{3} \frac{3}{u} d u

\int_{1}^{3} 1 d u = 2

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

= 2 + 3 ln (3)