integral of (x^3)/(x^2+4)

解答

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

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

求解步骤

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

使用换元积分法

= \int \frac{u - 4}{2 u} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 4}{u} d u

乘开

\frac{u - 4}{u} : 1 - \frac{4}{u}

= \frac{1}{2} \cdot \int 1 - \frac{4}{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

= \frac{1}{2} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{1}{2} (u - 4 ln ∣ u ∣)

u = x^{2} + 4

代回

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

解答补常数

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

l a pl a ce t r an s f or m 5 sin (3 t - \frac{π}{4})

a re a e^{x}, e^{- x}, [0, 1]

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x \to 0 + lim ((1 + x)^{(\frac{1}{x})})

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