integral of (1+2x)^{-2}

解答

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

- \frac{1}{2 ( 1 + 2 x )} + C

求解步骤

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

使用换元积分法

= \int \frac{1}{2 u ^{2}} d u

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

使用幂法则

= \frac{1}{2} (- \frac{1}{u})

u = 1 + 2 x

代回

= \frac{1}{2} (- \frac{1}{1 + 2 x})

化简

\frac{1}{2} (- \frac{1}{1 + 2 x}) : - \frac{1}{2 ( 1 + 2 x )}

= - \frac{1}{2 ( 1 + 2 x )}

解答补常数

= - \frac{1}{2 ( 1 + 2 x )} + C

y^{^{''}} e^{a x} - 2 a^{2} y e^{a x} = 0

t an g e n t f (x) = 3 x + 7, at x = 6

d er i v a t i v e \frac{1 - 3 x}{6 + x}

d er i v a t i v e f (x) = \frac{1}{( x + 1 ) ^{2}}

n = 1 \sum \infty \frac{n}{2 ^{n}}