integral of (3x-1)/((3x^2-2x+1)^4)

解答

\int \frac{3 x - 1}{( 3 x ^{2} - 2 x + 1 ) ^{4}} d x

- \frac{1}{6 ( 3 x ^{2} - 2 x + 1 ) ^{3}} + C

求解步骤

\int \frac{3 x - 1}{( 3 x ^{2} - 2 x + 1 ) ^{4}} d x

使用换元积分法

= \int \frac{1}{2 u ^{4}} 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 ^{4}} d u

使用幂法则

= \frac{1}{2} (- \frac{1}{3 u ^{3}})

u = 3 x^{2} - 2 x + 1

代回

= \frac{1}{2} (- \frac{1}{3 ( 3 x ^{2} - 2 x + 1 ) ^{3}})

化简

\frac{1}{2} (- \frac{1}{3 ( 3 x ^{2} - 2 x + 1 ) ^{3}}) : - \frac{1}{6 ( 3 x ^{2} - 2 x + 1 ) ^{3}}

= - \frac{1}{6 ( 3 x ^{2} - 2 x + 1 ) ^{3}}

解答补常数

= - \frac{1}{6 ( 3 x ^{2} - 2 x + 1 ) ^{3}} + C

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