integral of x*sqrt(1+4x^2)

解答

\int x \cdot 1 + 4 x^{2} d x

\frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}} + C

求解步骤

\int x 1 + 4 x^{2} d x

使用换元积分法

= \int \frac{u}{8} d u

提出常数:

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

= \frac{1}{8} \cdot \int u d u

使用幂法则

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

u = 1 + 4 x^{2}

代回

= \frac{1}{8} \cdot \frac{2}{3} (1 + 4 x^{2})^{\frac{3}{2}}

化简

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

= \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}}

解答补常数

= \frac{1}{12} (1 + 4 x^{2})^{\frac{3}{2}} + C