integral of x(5-2x)^{1/3}

解答

\int x (5 - 2 x)^{\frac{1}{3}} d x

\frac{1}{4} (\frac{3}{7} (5 - 2 x)^{\frac{7}{3}} - \frac{15}{4} (5 - 2 x)^{\frac{4}{3}}) + C

求解步骤

\int x (5 - 2 x)^{\frac{1}{3}} d x

使用换元积分法

= \int \frac{u ^{\frac{1}{3}} ( u - 5 )}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int u^{\frac{1}{3}} (u - 5) d u

乘开

u^{\frac{1}{3}} (u - 5) : u^{\frac{4}{3}} - 5 u^{\frac{1}{3}}

= \frac{1}{4} \cdot \int u^{\frac{4}{3}} - 5 u^{\frac{1}{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} (\int u^{\frac{4}{3}} d u - \int 5 u^{\frac{1}{3}} d u)

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

\int 5 u^{\frac{1}{3}} d u = \frac{15}{4} u^{\frac{4}{3}}

= \frac{1}{4} (\frac{3}{7} u^{\frac{7}{3}} - \frac{15}{4} u^{\frac{4}{3}})

u = 5 - 2 x

代回

= \frac{1}{4} (\frac{3}{7} (5 - 2 x)^{\frac{7}{3}} - \frac{15}{4} (5 - 2 x)^{\frac{4}{3}})

解答补常数

= \frac{1}{4} (\frac{3}{7} (5 - 2 x)^{\frac{7}{3}} - \frac{15}{4} (5 - 2 x)^{\frac{4}{3}}) + C

\int \frac{x ^{2}}{( a ^{2} + x ^{2} ) ^{2}} d x

in t e g r a l \frac{2 x}{9 x ^{2} - 6 x - 8}

\int 2 + cos^{2} (x) sin (x) d x

\int \frac{x}{2} sin (x^{2}) d x

\int \frac{x ^{3} - x}{x ^{2} + 4 x + 9} d x