integral of sqrt(1/(4x)+9)

解答

\int \frac{1}{4 x} + 9 d x

\frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1) + C

求解步骤

\int \frac{1}{4 x} + 9 d x

\frac{1}{4 x} + 9 = 9 (\frac{\frac{1}{4 x}}{9} + 1)

= \int 9 (\frac{\frac{1}{4 x}}{9} + 1) d x

化简

9 (\frac{\frac{1}{4 x}}{9} + 1) : \frac{1 + 36 x}{2 x}

= \int \frac{1 + 36 x}{2 x} d x

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1 + 36 x}{x} d x

使用分布积分法

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

\int \frac{36 x}{1 + 36 x} d x = \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)

= \frac{1}{2} (2 x 1 + 36 x - \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1))

化简

\frac{1}{2} (2 x 1 + 36 x - \frac{1}{6} (- ln 6 x + 36 x + 1 + 6 x 36 x + 1)) : \frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1)

= \frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1)

解答补常数

= \frac{1}{12} (6 x 36 x + 1 + ln 6 x + 36 x + 1) + C

y^{^{'}} = \frac{( x ^{4} + x ^{2} y ^{2} + y ^{4} )}{x ^{3} y}

l a pl a ce t r an s f or m t - 2

d er i v a t i v e y = lo g_{6} (x)

x \to 2 lim (x^{2})

\int \frac{1 + y}{y} d y