What is the integral of 1/(sqrt(8x)) ?

The integral of 1/(sqrt(8x)) is 1/(sqrt(2))sqrt(x)+C

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integral of 1/(sqrt(8x))

\int \frac{1}{8 x} d x

\frac{1}{2} x + C

Solution steps

\int \frac{1}{8 x} d x

Apply Integration By Parts

: \frac{1}{2 2} x - \int - \frac{1}{4 2 x ^{\frac{1}{2}}} d x

= \frac{1}{2 2} x - \int - \frac{1}{4 2 x ^{\frac{1}{2}}} d x

\int - \frac{1}{4 2 x ^{\frac{1}{2}}} d x = - \frac{1}{2 2} x

= \frac{1}{2 2} x - (- \frac{1}{2 2} x)

Simplify

\frac{1}{2 2} x - (- \frac{1}{2 2} x) : \frac{1}{2} x

= \frac{1}{2} x

Add a constant to the solution

= \frac{1}{2} x + C

\frac{d y}{d x} = \frac{x ^{6} - 2 y}{x}

\frac{\partial}{\partial y} (x y^{2} cos (2 + x^{2} y))

\frac{2}{( s - 2 )}

\int x (4 x + 1)^{2} d x

n = 1 \sum \infty (- 1)^{n} \frac{2 n !}{2 ^{n} n ! n}

What is the integral of 1/(sqrt(8x)) ?
The integral of 1/(sqrt(8x)) is 1/(sqrt(2))sqrt(x)+C