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受欢迎的微积分 >

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

解答

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

解答

ln x + 1 - ln x - 1 - 2 x + C

求解步骤

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

使用换元积分法

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

提出常数:

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

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

\frac{u ^{2}}{1 - u ^{2}} = \frac{1}{1 - u ^{2}} - 1

= 2 \cdot \int \frac{1}{1 - u ^{2}} - 1 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

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

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

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

u = x

代回

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

化简

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

= ln x + 1 - ln x - 1 - 2 x

解答补常数

= ln x + 1 - ln x - 1 - 2 x + C

作图

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\int_{- 1}^{0} 2 x - e^{x} d x

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derivative of 5(x^2+7^4(2x))

\frac{d}{d x} (5 (x^{2} + 7)^{4} (2 x))

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h \to 0 lim (\frac{( 1 + h ) ^{3} - 1}{h})

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\int \frac{1}{1 + cos ( x )} d x