integral from k to 2k of 2/((2x-k)^2)

解答

\int_{k}^{2 k} \frac{2}{( 2 x - k ) ^{2}} d x

\frac{2}{3 k}

求解步骤

\int_{k}^{2 k} \frac{2}{( 2 x - k ) ^{2}} d x

提出常数:

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

= 2 \cdot \int_{k}^{2 k} \frac{1}{( 2 x - k ) ^{2}} d x

使用换元积分法

= 2 \cdot \int_{k}^{3 k} \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

= 2 \cdot \frac{1}{2} \cdot \int_{k}^{3 k} \frac{1}{u ^{2}} d u

使用幂法则

= 2 \cdot \frac{1}{2} [- \frac{1}{u}]_{k}^{3 k}

化简

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{k}^{3 k} : [- \frac{1}{u}]_{k}^{3 k}

= [- \frac{1}{u}]_{k}^{3 k}

带入上下限计算后得:

- \frac{1}{3 k} + \frac{1}{k}

= - \frac{1}{3 k} + \frac{1}{k}

化简

= \frac{2}{3 k}

\int \frac{sec ( x )}{sec ( x ) - cos ( x )} d x

\frac{d}{d x} (- 18)

\int \frac{x}{2 x} d x

\frac{d}{d x} (x^{4} + 3 x^{3} - x^{2})

in v erse l a pl a ce \frac{s + 1}{( s ^{2} - 4 s ) ( s + 5 )}