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受欢迎的微积分 >

integral of 1/(x*sqrt(1+x^2))

解答

\int \frac{1}{x \cdot 1 + x ^{2}} d x

解答

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

求解步骤

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

使用换元积分法

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

因式分解

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

= \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

= - \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}

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

u = 1 + x^{2}

代回

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

化简

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

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

解答补常数

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

作图

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