integral of (3x+8)/(x^3+4x)

解答

\int \frac{3 x + 8}{x ^{3} + 4 x} d x

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

求解步骤

\int \frac{3 x + 8}{x ^{3} + 4 x} d x

乘开

\frac{3 x + 8}{x ^{3} + 4 x} : \frac{3 x}{x ^{3} + 4 x} + \frac{8}{x ^{3} + 4 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}{x ^{3} + 4 x} d x + \int \frac{8}{x ^{3} + 4 x} d x

\int \frac{3 x}{x ^{3} + 4 x} d x = \frac{3}{2} arctan (\frac{x}{2})

\int \frac{8}{x ^{3} + 4 x} d x = 2 ln ∣ x ∣ - ln x^{2} + 4

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

解答补常数

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

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