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

解答

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

- \frac{2}{x - 1} + ln ∣ x ∣ + C

求解步骤

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

乘开

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

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

\int \frac{x ^{2}}{x ( x - 1 ) ^{2}} d x = ln ∣ x - 1 ∣ - \frac{1}{x - 1}

\int \frac{1}{x ( x - 1 ) ^{2}} d x = ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1}

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

化简

ln ∣ x - 1 ∣ - \frac{1}{x - 1} + ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1} : - \frac{2}{x - 1} + ln ∣ x ∣

= - \frac{2}{x - 1} + ln ∣ x ∣

解答补常数

= - \frac{2}{x - 1} + ln ∣ x ∣ + C

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