integral of (2-x^2)/((x^3-6x+1)^5)

解答

\int \frac{2 - x ^{2}}{( x ^{3} - 6 x + 1 ) ^{5}} d x

\frac{1}{12 ( x ^{3} - 6 x + 1 ) ^{4}} + C

求解步骤

\int \frac{2 - x ^{2}}{( x ^{3} - 6 x + 1 ) ^{5}} d x

使用换元积分法

= \int - \frac{1}{3 u ^{5}} d u

提出常数:

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

= - \frac{1}{3} \cdot \int \frac{1}{u ^{5}} d u

使用幂法则

= - \frac{1}{3} (- \frac{1}{4 u ^{4}})

u = x^{3} - 6 x + 1

代回

= - \frac{1}{3} (- \frac{1}{4 ( x ^{3} - 6 x + 1 ) ^{4}})

化简

- \frac{1}{3} (- \frac{1}{4 ( x ^{3} - 6 x + 1 ) ^{4}}) : \frac{1}{12 ( x ^{3} - 6 x + 1 ) ^{4}}

= \frac{1}{12 ( x ^{3} - 6 x + 1 ) ^{4}}

解答补常数

= \frac{1}{12 ( x ^{3} - 6 x + 1 ) ^{4}} + C

\int (3 x - 1)^{4} d x

\int_{0}^{1} \frac{37}{x ^{p}} d x

\frac{d}{d x} (e^{3 x^{2}} \cdot 6 x)

t an g e n t y = 2 x^{3} + 3 x^{2} - 12 x + 6

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + y = 1