integral of (ax+b)^n

解答

\int (a x + b)^{n} d x

\frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + C

求解步骤

\int (a x + b)^{n} d x

使用换元积分法

= \int \frac{u ^{n}}{a} d u

提出常数:

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

= \frac{1}{a} \cdot \int u^{n} d u

使用幂法则

= \frac{1}{a} \cdot \frac{u ^{n + 1}}{n + 1}

u = a x + b

代回

= \frac{1}{a} \cdot \frac{( a x + b ) ^{n + 1}}{n + 1}

化简

\frac{1}{a} \cdot \frac{( a x + b ) ^{n + 1}}{n + 1} : \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )}

= \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )}

解答补常数

= \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + C

\frac{d}{d x} (e^{3 x} \cdot cos (2 x))

\int_{0}^{\frac{π}{12}} 6 tan (3 x) d x

x \to 2 lim (2 x^{3} - 3 x^{2} + 4 x - 9)

y^{^{''}} + 2 y^{^{'}} + y = x e^{- x}

x \to 0 lim (\frac{3}{x ( x + 3 ) ^{2}})