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

解答

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

\frac{1}{24}

十进制

0.04166 \dots

求解步骤

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

使用分布积分法

= [- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - \int - \frac{2 x}{3 ( x + 1 ) ^{3}} d x]_{0}^{1}

\int - \frac{2 x}{3 ( x + 1 ) ^{3}} d x = - \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}}

= [- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - (- \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}})]_{0}^{1}

化简

[- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - (- \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}})]_{0}^{1} : [\frac{- 3 x ^{2} - 3 x - 1}{3 ( x + 1 ) ^{3}}]_{0}^{1}

= [\frac{- 3 x ^{2} - 3 x - 1}{3 ( x + 1 ) ^{3}}]_{0}^{1}

带入上下限计算后得:

\frac{1}{24}

= \frac{1}{24}

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