integral of (8x+3)^4

解答

\int (8 x + 3)^{4} d x

\frac{1}{40} (8 x + 3)^{5} + C

求解步骤

\int (8 x + 3)^{4} d x

使用换元积分法

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

使用幂法则

= \frac{1}{8} \cdot \frac{u ^{5}}{5}

u = 8 x + 3

代回

= \frac{1}{8} \cdot \frac{( 8 x + 3 ) ^{5}}{5}

化简

\frac{1}{8} \cdot \frac{( 8 x + 3 ) ^{5}}{5} : \frac{1}{40} (8 x + 3)^{5}

= \frac{1}{40} (8 x + 3)^{5}

解答补常数

= \frac{1}{40} (8 x + 3)^{5} + C

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

a re a (48 - 2 x) - x, [- 8, 24]

x \to \infty lim (\frac{- 5}{x})

in v erse l a pl a ce \frac{1}{( s - 3 )}

y^{^{'}} = x 1 - y^{2}