What is the integral of 1/(2x^2y) ?

The integral of 1/(2x^2y) is -1/(2yx)+C

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integral of 1/(2x^2y)

integral

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

- \frac{1}{2 y x} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

Simplify

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

= - \frac{1}{2 y x}

Add a constant to the solution

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

in t e g r a l x^{4} e^{x}

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in t e g r a l ln (1 - x)

in t e g r a l 2 e^{- 2} x

in t e g r a l \frac{( x ^{3} - x ^{2} + x - 1 )}{( x - 1 )}