integral of 1/(4+sqrt(2x))

解答

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

4 + 2 x - 4 ln 4 + 2 x + C

求解步骤

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

使用换元积分法

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

乘开

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

= \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

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

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= u - 4 ln ∣ u ∣

u = 4 + 2 x

代回

= 4 + 2 x - 4 ln 4 + 2 x

化简

4 + 2 x - 4 ln 4 + 2 x : 4 + 2 x - 4 ln 4 + 2 x

= 4 + 2 x - 4 ln 4 + 2 x

解答补常数

= 4 + 2 x - 4 ln 4 + 2 x + C

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