What is the integral of y/(sqrt(x)) ?

The integral of y/(sqrt(x)) is 2ysqrt(x)+C

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integral of y/(sqrt(x))

\int \frac{y}{x} d x

2 y x + C

Solution steps

\int \frac{y}{x} d x

Take the constant out:

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

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

Apply the Power Rule

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

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

Apply radical rule:

= 2 y x

Add a constant to the solution

= 2 y x + C

y = (x^{3} - 16 x)^{8}, (- 4, 0)

\frac{d y}{d x} = (x + y)

\frac{d}{d x} (x (ln (x^{2})))

\int sin^{6} (2 x) d x

x^{^{''}} + 3 x^{^{'}} + 4 x = 0, x (0) = 3, x^{^{'}} (0) = 1

What is the integral of y/(sqrt(x)) ?
The integral of y/(sqrt(x)) is 2ysqrt(x)+C