integral of x/(sqrt(16x^2-1))

解答

\int \frac{x}{16 x ^{2} - 1} d x

\frac{1}{16} 16 x^{2} - 1 + C

求解步骤

\int \frac{x}{16 x ^{2} - 1} d x

使用换元积分法

= \int \frac{1}{32 u} d u

提出常数:

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

= \frac{1}{32} \cdot \int \frac{1}{u} d u

使用幂法则

= \frac{1}{32} \cdot 2 u^{\frac{1}{2}}

u = 16 x^{2} - 1

代回

= \frac{1}{32} \cdot 2 (16 x^{2} - 1)^{\frac{1}{2}}

化简

\frac{1}{32} \cdot 2 (16 x^{2} - 1)^{\frac{1}{2}} : \frac{1}{16} 16 x^{2} - 1

= \frac{1}{16} 16 x^{2} - 1

解答补常数

= \frac{1}{16} 16 x^{2} - 1 + C

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