integral from 0 to 1 of (x^4)/(x^2+1)

解答

\int_{0}^{1} \frac{x ^{4}}{x ^{2} + 1} d x

- \frac{2}{3} + \frac{π}{4}

十进制

0.11873 \dots

求解步骤

\int_{0}^{1} \frac{x ^{4}}{x ^{2} + 1} d x

长除

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

= \int_{0}^{1} x^{2} - 1 + \frac{1}{x ^{2} + 1} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} x^{2} d x - \int_{0}^{1} 1 d x + \int_{0}^{1} \frac{1}{x ^{2} + 1} d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} \frac{1}{x ^{2} + 1} d x = \frac{π}{4}

= \frac{1}{3} - 1 + \frac{π}{4}

化简

\frac{1}{3} - 1 + \frac{π}{4} : - \frac{2}{3} + \frac{π}{4}

= - \frac{2}{3} + \frac{π}{4}

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