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

解答

\int \frac{- 2 \cdot x ^{4}}{2 \cdot x ^{3} - 3 \cdot x ^{2}} d x

- 2 (\frac{x ^{2}}{4} + \frac{3}{4} x + \frac{9}{8} ln ∣ 2 x - 3 ∣) + C

求解步骤

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

化简

\frac{x ^{4}}{2 x ^{3} - 3 x ^{2}} : \frac{x ^{2}}{2 x - 3}

= - 2 \cdot \int \frac{x ^{2}}{2 x - 3} d x

长除

\frac{x ^{2}}{2 x - 3} : \frac{x}{2} + \frac{3}{4} + \frac{9}{4 ( 2 x - 3 )}

= - 2 \cdot \int \frac{x}{2} + \frac{3}{4} + \frac{9}{4 ( 2 x - 3 )} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 2 (\int \frac{x}{2} d x + \int \frac{3}{4} d x + \int \frac{9}{4 ( 2 x - 3 )} d x)

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

\int \frac{3}{4} d x = \frac{3}{4} x

\int \frac{9}{4 ( 2 x - 3 )} d x = \frac{9}{8} ln ∣ 2 x - 3 ∣

= - 2 (\frac{x ^{2}}{4} + \frac{3}{4} x + \frac{9}{8} ln ∣ 2 x - 3 ∣)

解答补常数

= - 2 (\frac{x ^{2}}{4} + \frac{3}{4} x + \frac{9}{8} ln ∣ 2 x - 3 ∣) + C

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