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

The integral of 1/(8sqrt(x)) is 1/4 sqrt(x)+C

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integral of 1/(8sqrt(x))

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

\frac{1}{4} x + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: 2 x^{\frac{1}{2}}

= \frac{1}{8} \cdot 2 x^{\frac{1}{2}}

Simplify

\frac{1}{8} \cdot 2 x^{\frac{1}{2}} : \frac{1}{4} x

= \frac{1}{4} x

Add a constant to the solution

= \frac{1}{4} x + C

y^{^{''}} - 5 y^{^{'}} + 6 y = 36 t

x \frac{d}{d x} [ln (y)] = 1

y = x, y = 4 x, y = - x + 2

\int 6 sin (3 x) d x

\frac{x ^{2} - 3}{( x - 1 ) ^{2} + ( x - 1 ) ^{2}}

What is the integral of 1/(8sqrt(x)) ?
The integral of 1/(8sqrt(x)) is 1/4 sqrt(x)+C