integral of sqrt(1+(1/(2\sqrt{x)))^2}

解答

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

- \frac{1}{8} (- 4 x 4 x + 1 - ln \frac{4 x + 1}{2 x} + 1 + ln \frac{4 x + 1}{2 x} - 1) + C

求解步骤

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

使用换元积分法

= \int - \frac{1 + u}{4 u u ^{\frac{3}{2}}} d u

提出常数:

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

= - \frac{1}{4} \cdot \int \frac{1 + u}{u u ^{\frac{3}{2}}} d u

u^{\frac{3}{2}} = u u

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

u u u = u^{2}

= - \frac{1}{4} \cdot \int \frac{1 + u}{u ^{2}} d u

使用分布积分法

= - \frac{1}{4} (- \frac{1 + u}{u} - \int - \frac{1}{2 u 1 + u} d u)

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

= - \frac{1}{4} (- \frac{1 + u}{u} - (\frac{1}{2} ln 1 + u + 1 - \frac{1}{2} ln 1 + u - 1))

u = (\frac{1}{2 x})^{2}

代回

= - \frac{1}{4} - \frac{1 + ( \frac{1}{2 x} ) ^{2}}{( \frac{1}{2 x} ) ^{2}} - \frac{1}{2} ln 1 + (\frac{1}{2 x})^{2} + 1 - \frac{1}{2} ln 1 + (\frac{1}{2 x})^{2} - 1

化简

- \frac{1}{4} - \frac{1 + ( \frac{1}{2 x} ) ^{2}}{( \frac{1}{2 x} ) ^{2}} - \frac{1}{2} ln 1 + (\frac{1}{2 x})^{2} + 1 - \frac{1}{2} ln 1 + (\frac{1}{2 x})^{2} - 1 : - \frac{1}{8} (- 4 x 4 x + 1 - ln \frac{4 x + 1}{2 x} + 1 + ln \frac{4 x + 1}{2 x} - 1)

= - \frac{1}{8} (- 4 x 4 x + 1 - ln \frac{4 x + 1}{2 x} + 1 + ln \frac{4 x + 1}{2 x} - 1)

解答补常数

= - \frac{1}{8} (- 4 x 4 x + 1 - ln \frac{4 x + 1}{2 x} + 1 + ln \frac{4 x + 1}{2 x} - 1) + C

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