What is the integral of ((2y^{1/2})/(y^3))0x ?

The integral of ((2y^{1/2})/(y^3))0x is -(4x)/(3y^{3/2)}+C

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integral of ((2y^{1/2})/(y^3))0x

\int (\frac{2 y ^{\frac{1}{2}}}{y ^{3}}) d^{0} x

- \frac{4 x}{3 y ^{\frac{3}{2}}} + C

Solution steps

\int \frac{2 y ^{\frac{1}{2}}}{y ^{3}} d^{0} x d y

Take the constant out:

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

= 2 d^{0} x \cdot \int \frac{y ^{\frac{1}{2}}}{y ^{3}} d y

Simplify

= 2 x \cdot \int \frac{y ^{\frac{1}{2}}}{y ^{3}} d y

Simplify

\frac{y ^{\frac{1}{2}}}{y ^{3}} : \frac{1}{y ^{\frac{5}{2}}}

= 2 x \cdot \int \frac{1}{y ^{\frac{5}{2}}} d y

Apply the Power Rule

: - \frac{2}{3 y ^{\frac{3}{2}}}

= 2 x (- \frac{2}{3 y ^{\frac{3}{2}}})

Simplify

2 x (- \frac{2}{3 y ^{\frac{3}{2}}}) : - \frac{4 x}{3 y ^{\frac{3}{2}}}

= - \frac{4 x}{3 y ^{\frac{3}{2}}}

Add a constant to the solution

= - \frac{4 x}{3 y ^{\frac{3}{2}}} + C

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

(1 + ln (x) + \frac{y}{x}) d x + (- 1 + ln (x)) d y = 0

y = e^{3 x} + 10 e^{2 x}

x \to 0 lim (\frac{2 e ^{x} - 2}{3 x})

t e^{\frac{- t}{2}}

What is the integral of ((2y^{1/2})/(y^3))0x ?
The integral of ((2y^{1/2})/(y^3))0x is -(4x)/(3y^{3/2)}+C