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

解答

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

3 ln ∣ x ∣ + \frac{4}{x} - ln ∣ x + 1 ∣ + C

求解步骤

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

将

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

用部份分式展开:

\frac{3}{x} - \frac{4}{x ^{2}} - \frac{1}{x + 1}

= \int \frac{3}{x} - \frac{4}{x ^{2}} - \frac{1}{x + 1} 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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

= 3 ln ∣ x ∣ - (- \frac{4}{x}) - ln ∣ x + 1 ∣

化简

= 3 ln ∣ x ∣ + \frac{4}{x} - ln ∣ x + 1 ∣

解答补常数

= 3 ln ∣ x ∣ + \frac{4}{x} - ln ∣ x + 1 ∣ + C

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