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

解

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

8 ln (2) - \frac{3}{2} - 4 ln (3)

十進法表記

- 0.34927 \dots

解答ステップ

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

以下の部分分数を得る:

\frac{x - 3}{x ^{3} + x ^{2}} : \frac{4}{x} - \frac{3}{x ^{2}} - \frac{4}{x + 1}

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

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

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

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

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

簡素化

4 ln (2) - \frac{3}{2} - 4 (ln (3) - ln (2)) : 8 ln (2) - \frac{3}{2} - 4 ln (3)

= 8 ln (2) - \frac{3}{2} - 4 ln (3)