integral of sqrt(1+1/(4x))

解答

\int 1 + \frac{1}{4 x} d x

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

求解步骤

\int 1 + \frac{1}{4 x} d x

使用换元积分法

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

使用分布积分法

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

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

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

u = 4 x

代回

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

化简

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

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

解答补常数

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

\frac{d}{d x} (cot (3 x^{4} - 2 x^{3} - 3))

x \to 4 lim (\frac{6 x - 1}{2 x - 9})

\int - 2.1 x d x

\int_{0}^{4} (4 x + 7) d x

\frac{\partial}{\partial y} (2 y e^{2 z})