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受欢迎的微积分 >

integral of x^3sqrt(a^2x^2+b^2)

解答

\int x^{3} a^{2} x^{2} + b^{2} d x

解答

\frac{( 3 a ^{2} x ^{2} - 2 b ^{2} ) ( a ^{2} x ^{2} + b ^{2} ) ^{\frac{3}{2}}}{15 a ^{4}} + C

求解步骤

\int x^{3} a^{2} x^{2} + b^{2} d x

化简

x^{3} : x^{2} x

= \int x^{2} x a^{2} x^{2} + b^{2} d x

使用换元积分法

= \int \frac{u a ^{2} u + b ^{2}}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int u a^{2} u + b^{2} d u

使用换元积分法

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

提出常数:

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

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

乘开

v (v - b^{2}) : v^{\frac{3}{2}} - b^{2} v

= \frac{1}{2} \cdot \frac{1}{a ^{4}} \cdot \int v^{\frac{3}{2}} - b^{2} v d v

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

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

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int b^{2} v d v = b^{2} \frac{2}{3} v^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{1}{a ^{4}} (\frac{2}{5} v^{\frac{5}{2}} - b^{2} \frac{2}{3} v^{\frac{3}{2}})

代回

= \frac{1}{2} \cdot \frac{1}{a ^{4}} (\frac{2}{5} (a^{2} x^{2} + b^{2})^{\frac{5}{2}} - b^{2} \frac{2}{3} (a^{2} x^{2} + b^{2})^{\frac{3}{2}})

化简

\frac{1}{2} \cdot \frac{1}{a ^{4}} (\frac{2}{5} (a^{2} x^{2} + b^{2})^{\frac{5}{2}} - b^{2} \frac{2}{3} (a^{2} x^{2} + b^{2})^{\frac{3}{2}}) : \frac{( 3 a ^{2} x ^{2} - 2 b ^{2} ) ( a ^{2} x ^{2} + b ^{2} ) ^{\frac{3}{2}}}{15 a ^{4}}

= \frac{( 3 a ^{2} x ^{2} - 2 b ^{2} ) ( a ^{2} x ^{2} + b ^{2} ) ^{\frac{3}{2}}}{15 a ^{4}}

解答补常数

= \frac{( 3 a ^{2} x ^{2} - 2 b ^{2} ) ( a ^{2} x ^{2} + b ^{2} ) ^{\frac{3}{2}}}{15 a ^{4}} + C

作图

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