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受欢迎的微积分 >

integral of (-x)/(1+x)

解答

\int \frac{- x}{1 + x} d x

解答

- 1 - x + ln ∣ 1 + x ∣ + C

求解步骤

\int \frac{- x}{1 + x} d x

提出常数:

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

= - \int \frac{x}{1 + x} d x

使用换元积分法

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

乘开

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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

u = 1 + x

代回

= - (1 + x - ln ∣ 1 + x ∣)

化简

= - 1 - x + ln ∣ 1 + x ∣

解答补常数

= - 1 - x + ln ∣ 1 + x ∣ + C

作图

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derivative of sqrt(x)e^{-x^2}

\frac{d}{d x} (x e^{- x^{2}})

f(x)=10^{sqrt(x)}

f (x) = 1 0^{x}

derivative (2x+1)/(x+2)

d er i v a t i v e \frac{2 x + 1}{x + 2}

y^{^{'}} = (2 t + 1) e^{y}

d/(dz)(sqrt(z+1)-1)

\frac{d}{d z} (z + 1 - 1)