integral of x\sqrt[3]{4x+1}

解答

\int x 3 4 x + 1 d x

\frac{1}{16} (\frac{3}{7} (4 x + 1)^{\frac{7}{3}} - \frac{3}{4} (4 x + 1)^{\frac{4}{3}}) + C

求解步骤

\int x 3 4 x + 1 d x

使用换元积分法

= \int \frac{3 u ( u - 1 )}{16} d u

提出常数:

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

= \frac{1}{16} \cdot \int 3 u (u - 1) d u

乘开

3 u (u - 1) : u^{\frac{4}{3}} - 3 u

= \frac{1}{16} \cdot \int u^{\frac{4}{3}} - 3 u 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}{16} (\int u^{\frac{4}{3}} d u - \int 3 u d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

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

= \frac{1}{16} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

u = 4 x + 1

代回

= \frac{1}{16} (\frac{3}{7} (4 x + 1)^{\frac{7}{3}} - \frac{3}{4} (4 x + 1)^{\frac{4}{3}})

解答补常数

= \frac{1}{16} (\frac{3}{7} (4 x + 1)^{\frac{7}{3}} - \frac{3}{4} (4 x + 1)^{\frac{4}{3}}) + C

d er i v a t i v e \frac{3 ( 7 - x ) + 5 9 + x ^{2}}{15}

d er i v a t i v e \frac{x ^{2} - 2}{2 x + 1}

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

x \to 0 lim (\frac{1}{x} - \frac{1}{x ^{2} + x})

d er i v a t i v e f (x) = 4 + 3 x