What is the integral of-(((y^2+1))/((2xy))) ?

The integral of-(((y^2+1))/((2xy))) is -1/(2x)(ln|y|+(y^2)/2)+C

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integral of-(((y^2+1))/((2xy)))

integral

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

- \frac{1}{2 x} (ln ∣ y ∣ + \frac{y ^{2}}{2}) + C

Solution steps

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

Take the constant out:

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

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

Expand

\frac{1 + y ^{2}}{y} : \frac{1}{y} + y

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{y} d y = ln ∣ y ∣

\int y d y = \frac{y ^{2}}{2}

= - \frac{1}{2 x} (ln ∣ y ∣ + \frac{y ^{2}}{2})

Add a constant to the solution

= - \frac{1}{2 x} (ln ∣ y ∣ + \frac{y ^{2}}{2}) + C

\frac{2}{s ^{2}}

\frac{1}{( y - y ^{2} )}

\frac{1}{( x \cdot ( e ^{2} ) ^{x} )}

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

\frac{sin ^{4} ( x )}{( 1 + cos ( x ) ) ^{2}}

What is the integral of-(((y^2+1))/((2xy))) ?
The integral of-(((y^2+1))/((2xy))) is -1/(2x)(ln|y|+(y^2)/2)+C