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受欢迎的微积分 >

integral of sqrt(1+(x^3-1/(4x^3))^2)

解答

\int 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

解答

\frac{2 x ^{6} - 1}{8 x ^{2}} + C

求解步骤

\int 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

化简

1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} : \frac{8 x ^{6} + 1 + 16 x ^{12}}{4 x ^{3}}

= \int \frac{8 x ^{6} + 1 + 16 x ^{12}}{4 x ^{3}} d x

提出常数:

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

= \frac{1}{4} \cdot \int \frac{8 x ^{6} + 1 + 16 x ^{12}}{x ^{3}} d x

使用换元积分法

= \frac{1}{4} \cdot \int \frac{4 u + 1}{6 u ^{\frac{4}{3}}} d u

提出常数:

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

= \frac{1}{4} \cdot \frac{1}{6} \cdot \int \frac{4 u + 1}{u ^{\frac{4}{3}}} d u

乘开

\frac{4 u + 1}{u ^{\frac{4}{3}}} : \frac{4}{u ^{\frac{1}{3}}} + \frac{1}{u ^{\frac{4}{3}}}

= \frac{1}{4} \cdot \frac{1}{6} \cdot \int \frac{4}{u ^{\frac{1}{3}}} + \frac{1}{u ^{\frac{4}{3}}} 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

= \frac{1}{4} \cdot \frac{1}{6} (\int \frac{4}{u ^{\frac{1}{3}}} d u + \int \frac{1}{u ^{\frac{4}{3}}} d u)

\int \frac{4}{u ^{\frac{1}{3}}} d u = 6 u^{\frac{2}{3}}

\int \frac{1}{u ^{\frac{4}{3}}} d u = - \frac{3}{3 u}

= \frac{1}{4} \cdot \frac{1}{6} (6 u^{\frac{2}{3}} - \frac{3}{3 u})

u = x^{6}

代回

= \frac{1}{4} \cdot \frac{1}{6} (6 (x^{6})^{\frac{2}{3}} - \frac{3}{3 x ^{6}})

化简

\frac{1}{4} \cdot \frac{1}{6} (6 (x^{6})^{\frac{2}{3}} - \frac{3}{3 x ^{6}}) : \frac{2 x ^{6} - 1}{8 x ^{2}}

= \frac{2 x ^{6} - 1}{8 x ^{2}}

解答补常数

= \frac{2 x ^{6} - 1}{8 x ^{2}} + C

作图

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