integral from 0 to 1 of ((1-x)/(1+x))^3

解答

\int_{0}^{1} (\frac{1 - x}{1 + x})^{3} d x

- 13 + \frac{96 ln ( 2 ) + 144}{16}

十进制

0.15888 \dots

求解步骤

\int_{0}^{1} (\frac{1 - x}{1 + x})^{3} d x

乘开

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

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

乘开

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

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

\int_{0}^{1} \frac{1}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = \frac{3}{8}

\int_{0}^{1} \frac{3 x}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = \frac{3}{8}

\int_{0}^{1} \frac{3 x ^{2}}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2})

\int_{0}^{1} \frac{x ^{3}}{1 + 3 x + 3 x ^{2} + x ^{3}} d x = \frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2}

= \frac{3}{8} - \frac{3}{8} + 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}) - (\frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2})

化简

\frac{3}{8} - \frac{3}{8} + 3 (\frac{16 ln ( 2 ) + 30}{16} - \frac{5}{2}) - (\frac{- 48 ln ( 2 ) - 54}{16} + \frac{11}{2}) : - 13 + \frac{96 ln ( 2 ) + 144}{16}

= - 13 + \frac{96 ln ( 2 ) + 144}{16}

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