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受欢迎的微积分 >

integral of 1/(x^2sqrt(2x^2-32))

解答

\int \frac{1}{x ^{2} 2 x ^{2} - 32} d x

解答

\frac{x ^{2} - 16}{16 2 x} + C

求解步骤

\int \frac{1}{x ^{2} 2 x ^{2} - 32} d x

使用三角换元法

= \int \frac{1}{16 2 sec ( u )} d u

提出常数:

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

= \frac{1}{16 2} \cdot \int \frac{1}{sec ( u )} d u

使用三角恒等式改写

= \frac{1}{16 2} \cdot \int cos (u) d u

使用积分常见定则:

\int cos (u) d u = sin (u)

= \frac{1}{16 2} sin (u)

u = arcsec (\frac{2}{4 2} x)

代回

= \frac{1}{16 2} sin (arcsec (\frac{2}{4 2} x))

化简

\frac{1}{16 2} sin (arcsec (\frac{2}{4 2} x)) : \frac{x ^{2} - 16}{16 2 x}

= \frac{x ^{2} - 16}{16 2 x}

解答补常数

= \frac{x ^{2} - 16}{16 2 x} + C

作图

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