integral of (5x+1)^4

解答

\int (5 x + 1)^{4} d x

\frac{1}{25} (5 x + 1)^{5} + C

求解步骤

\int (5 x + 1)^{4} d x

使用换元积分法

= \int \frac{u ^{4}}{5} d u

提出常数:

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

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

使用幂法则

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

u = 5 x + 1

代回

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

化简

\frac{1}{5} \cdot \frac{( 5 x + 1 ) ^{5}}{5} : \frac{1}{25} (5 x + 1)^{5}

= \frac{1}{25} (5 x + 1)^{5}

解答补常数

= \frac{1}{25} (5 x + 1)^{5} + C

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{2} + 1})

d er i v a t i v e f (x) = 9 - ∣ x - 4 ∣ at x

x \to 5 - lim (\frac{2}{x ^{2} - 25})

\int \frac{10}{( x + 1 ) ( x ^{2} + 9 )} d x

s im pl i f y (e^{- x} - e^{2}) (e^{x} + e^{- 5})