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

解

\int_{2}^{e} \frac{x ^{2} + 3}{x ^{3} + 2 x} d x

\frac{3}{2} (1 - ln (2)) - \frac{1}{4} (ln (e^{2} + 2) - ln (6))

十進法表記

0.34833 \dots

解答ステップ

\int_{2}^{e} \frac{x ^{2} + 3}{x ^{3} + 2 x} d x

以下の部分分数を得る:

\frac{x ^{2} + 3}{x ^{3} + 2 x} : \frac{3}{2 x} - \frac{x}{2 ( x ^{2} + 2 )}

= \int_{2}^{e} \frac{3}{2 x} - \frac{x}{2 ( x ^{2} + 2 )} 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_{2}^{e} \frac{3}{2 x} d x - \int_{2}^{e} \frac{x}{2 ( x ^{2} + 2 )} d x

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

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

= \frac{3}{2} (1 - ln (2)) - \frac{1}{4} (ln (e^{2} + 2) - ln (6))