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受欢迎的微积分 >

integral of x^2sqrt(x^2-64)

解答

\int x^{2} x^{2} - 64 d x

解答

4096 (\frac{1}{12288} x^{3} x^{2} - 64 - \frac{1}{49152} x^{3} x^{2} - 64 - \frac{1}{512} x x^{2} - 64 - \frac{1}{8} ln (\frac{1}{8} x^{2} - 64 + x)) + C

求解步骤

\int x^{2} x^{2} - 64 d x

乘以

x^{2} - 64

的共轭根式

: \frac{x ^{4} - 4096}{x ^{2} + 64}

= \int x^{2} \frac{x ^{4} - 4096}{x ^{2} + 64} d x

使用换元积分法

= \int 4096 u^{2} (u + 1) (u - 1) d u

提出常数:

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

= 4096 \cdot \int u^{2} (u + 1) (u - 1) d u

乘以

u + 1

的共轭根式

: \frac{u ^{2} - 1}{u - 1}

= 4096 \cdot \int u^{2} \frac{u ^{2} - 1}{u - 1} (u - 1) d u

乘

\frac{u ^{2} - 1}{u - 1} (u - 1) : u^{2} - 1

= 4096 \cdot \int u^{2} u^{2} - 1 d u

使用分布积分法

= 4096 (\frac{1}{3} u^{3} u^{2} - 1 - \int \frac{u ^{4}}{3 u ^{2} - 1} d u)

\int \frac{u ^{4}}{3 u ^{2} - 1} d u = \frac{1}{12} u^{3} u^{2} - 1 + \frac{1}{8} (u u^{2} - 1 + ln u^{2} - 1 + u)

= 4096 (\frac{1}{3} u^{3} u^{2} - 1 - (\frac{1}{12} u^{3} u^{2} - 1 + \frac{1}{8} (u u^{2} - 1 + ln u^{2} - 1 + u)))

u = \frac{x}{8}

代回

= 4096 (\frac{1}{3} (\frac{x}{8})^{3} (\frac{x}{8})^{2} - 1 - (\frac{1}{12} (\frac{x}{8})^{3} (\frac{x}{8})^{2} - 1 + \frac{1}{8} (\frac{x}{8} (\frac{x}{8})^{2} - 1 + ln (\frac{x}{8})^{2} - 1 + \frac{x}{8})))

化简

4096 (\frac{1}{3} (\frac{x}{8})^{3} (\frac{x}{8})^{2} - 1 - (\frac{1}{12} (\frac{x}{8})^{3} (\frac{x}{8})^{2} - 1 + \frac{1}{8} (\frac{x}{8} (\frac{x}{8})^{2} - 1 + ln (\frac{x}{8})^{2} - 1 + \frac{x}{8}))) : 4096 (\frac{1}{12288} x^{3} x^{2} - 64 - \frac{1}{49152} x^{3} x^{2} - 64 - \frac{1}{512} x x^{2} - 64 - \frac{1}{8} ln (\frac{1}{8} x^{2} - 64 + x))

= 4096 (\frac{1}{12288} x^{3} x^{2} - 64 - \frac{1}{49152} x^{3} x^{2} - 64 - \frac{1}{512} x x^{2} - 64 - \frac{1}{8} ln (\frac{1}{8} x^{2} - 64 + x))

解答补常数

= 4096 (\frac{1}{12288} x^{3} x^{2} - 64 - \frac{1}{49152} x^{3} x^{2} - 64 - \frac{1}{512} x x^{2} - 64 - \frac{1}{8} ln (\frac{1}{8} x^{2} - 64 + x)) + C

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