integral from-1 to 1 of (x^4+2)/(x^2+1)

解

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

- \frac{4}{3} + \frac{π}{2} + π

十進法表記

3.37905 \dots

解答ステップ

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

拡張

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

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

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

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

= - \frac{4}{3} + \frac{π}{2} + π