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受欢迎的微积分 >

integral of x^4sqrt(x+1)

解答

\int x^{4} x + 1 d x

解答

2 (\frac{1}{11} (x + 1)^{\frac{11}{2}} - \frac{4}{9} (x + 1)^{\frac{9}{2}} + \frac{6}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{1}{3} (x + 1)^{\frac{3}{2}}) + C

求解步骤

\int x^{4} x + 1 d x

使用换元积分法

= \int 2 (u^{2} - 1)^{4} u^{2} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (u^{2} - 1)^{4} u^{2} d u

乘开

(u^{2} - 1)^{4} u^{2} : u^{10} - 4 u^{8} + 6 u^{6} - 4 u^{4} + u^{2}

= 2 \cdot \int u^{10} - 4 u^{8} + 6 u^{6} - 4 u^{4} + u^{2} d u

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

= 2 (\int u^{10} d u - \int 4 u^{8} d u + \int 6 u^{6} d u - \int 4 u^{4} d u + \int u^{2} d u)

\int u^{10} d u = \frac{u ^{11}}{11}

\int 4 u^{8} d u = \frac{4 u ^{9}}{9}

\int 6 u^{6} d u = \frac{6 u ^{7}}{7}

\int 4 u^{4} d u = \frac{4 u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

= 2 (\frac{u ^{11}}{11} - \frac{4 u ^{9}}{9} + \frac{6 u ^{7}}{7} - \frac{4 u ^{5}}{5} + \frac{u ^{3}}{3})

u = x + 1

代回

= 2 (\frac{( x + 1 ) ^{11}}{11} - \frac{4 ( x + 1 ) ^{9}}{9} + \frac{6 ( x + 1 ) ^{7}}{7} - \frac{4 ( x + 1 ) ^{5}}{5} + \frac{( x + 1 ) ^{3}}{3})

化简

2 (\frac{( x + 1 ) ^{11}}{11} - \frac{4 ( x + 1 ) ^{9}}{9} + \frac{6 ( x + 1 ) ^{7}}{7} - \frac{4 ( x + 1 ) ^{5}}{5} + \frac{( x + 1 ) ^{3}}{3}) : 2 (\frac{1}{11} (x + 1)^{\frac{11}{2}} - \frac{4}{9} (x + 1)^{\frac{9}{2}} + \frac{6}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{1}{3} (x + 1)^{\frac{3}{2}})

= 2 (\frac{1}{11} (x + 1)^{\frac{11}{2}} - \frac{4}{9} (x + 1)^{\frac{9}{2}} + \frac{6}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{1}{3} (x + 1)^{\frac{3}{2}})

解答补常数

= 2 (\frac{1}{11} (x + 1)^{\frac{11}{2}} - \frac{4}{9} (x + 1)^{\frac{9}{2}} + \frac{6}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{1}{3} (x + 1)^{\frac{3}{2}}) + C

作图

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