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

解答

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

- \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10 + C

求解步骤

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

提出常数:

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

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

使用分布积分法

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

\int - \frac{2 x ^{3}}{( x - 1 ) ^{2}} d x = - x^{2} - 4 x + \frac{2}{x - 1} - 6 ln ∣ x - 1 ∣ + 5

= 2 (- \frac{x ^{4}}{2 ( x - 1 ) ^{2}} - (- x^{2} - 4 x + \frac{2}{x - 1} - 6 ln ∣ x - 1 ∣ + 5))

化简

2 (- \frac{x ^{4}}{2 ( x - 1 ) ^{2}} - (- x^{2} - 4 x + \frac{2}{x - 1} - 6 ln ∣ x - 1 ∣ + 5)) : - \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10

= - \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10

解答补常数

= - \frac{x ^{4}}{( x - 1 ) ^{2}} + 2 x^{2} + 8 x - \frac{4}{x - 1} + 12 ln ∣ x - 1 ∣ - 10 + C

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