integral of (2/x+1)^2

解答

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

- \frac{4}{x} + 4 ln ∣ x ∣ + x + C

求解步骤

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

乘开

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

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

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int 1 d x = x

= - \frac{4}{x} + 4 ln ∣ x ∣ + x

解答补常数

= - \frac{4}{x} + 4 ln ∣ x ∣ + x + C

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