integral of xsqrt(a+bx)

解答

\int x a + b x d x

\frac{2 ( 5 b x ( a + b x ) ^{\frac{3}{2}} - 2 ( a + b x ) ^{\frac{5}{2}} )}{15 b ^{2}} + C

求解步骤

\int x a + b x d x

使用换元积分法

= \int \frac{2 u ^{2} ( u ^{2} - a )}{b ^{2}} d u

提出常数:

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

= \frac{2}{b ^{2}} \cdot \int u^{2} (u^{2} - a) d u

使用分布积分法

= \frac{2}{b ^{2}} (\frac{1}{3} u^{3} (u^{2} - a) - \int \frac{2 u ^{4}}{3} d u)

\int \frac{2 u ^{4}}{3} d u = \frac{2 u ^{5}}{15}

= \frac{2}{b ^{2}} (\frac{1}{3} u^{3} (u^{2} - a) - \frac{2 u ^{5}}{15})

u = a + b x

代回

= \frac{2}{b ^{2}} (\frac{1}{3} (a + b x)^{3} ((a + b x)^{2} - a) - \frac{2 ( a + b x ) ^{5}}{15})

化简

\frac{2}{b ^{2}} (\frac{1}{3} (a + b x)^{3} ((a + b x)^{2} - a) - \frac{2 ( a + b x ) ^{5}}{15}) : \frac{2 ( 5 b x ( a + b x ) ^{\frac{3}{2}} - 2 ( a + b x ) ^{\frac{5}{2}} )}{15 b ^{2}}

= \frac{2 ( 5 b x ( a + b x ) ^{\frac{3}{2}} - 2 ( a + b x ) ^{\frac{5}{2}} )}{15 b ^{2}}

解答补常数

= \frac{2 ( 5 b x ( a + b x ) ^{\frac{3}{2}} - 2 ( a + b x ) ^{\frac{5}{2}} )}{15 b ^{2}} + C

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