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受欢迎的微积分 >

integral of (18)/(xsqrt(1+2x))

解答

\int \frac{18}{x 1 + 2 x} d x

解答

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

求解步骤

\int \frac{18}{x 1 + 2 x} d x

提出常数:

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

= 18 \cdot \int \frac{1}{x 1 + 2 x} d x

使用换元积分法

= 18 \cdot \int \frac{2}{u ^{2} - 1} d u

提出常数:

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

= 18 \cdot 2 \cdot \int \frac{1}{u ^{2} - 1} d u

因式分解

u^{2} - 1 : - (- u^{2} + 1)

= 18 \cdot 2 \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

提出常数:

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

= 18 \cdot 2 (- \int \frac{1}{- u ^{2} + 1} d u)

使用积分常见定则:

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

= 18 \cdot 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

u = 1 + 2 x

代回

= 18 \cdot 2 (- (\frac{ln 1 + 2 x + 1}{2} - \frac{ln 1 + 2 x - 1}{2}))

化简

18 \cdot 2 (- (\frac{ln 1 + 2 x + 1}{2} - \frac{ln 1 + 2 x - 1}{2})) : - 36 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1)

= - 36 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1)

解答补常数

= - 36 (\frac{1}{2} ln 1 + 2 x + 1 - \frac{1}{2} ln 1 + 2 x - 1) + C

作图

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