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

解答

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

ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x}) + C

求解步骤

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

使用换元积分法

= \int - \frac{1}{u} - 1 - tanh^{2} (u) d u

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{1}{u} d u - \int 1 d u - \int tanh^{2} (u) d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int 1 d u = u

\int tanh^{2} (u) d u = u - tanh (u)

= - ln ∣ u ∣ - u - (u - tanh (u))

u = \frac{1}{x}

代回

= - ln \frac{1}{x} - \frac{1}{x} - (\frac{1}{x} - tanh (\frac{1}{x}))

化简

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

= ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x})

解答补常数

= ln ∣ x ∣ - \frac{2}{x} + tanh (\frac{1}{x}) + C

\int \frac{1}{1 + 3 x + 2} d x

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