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

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

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

\int \frac{( 1 + y ^{2} )}{y} d y

ln ∣ y ∣ + \frac{y ^{2}}{2} + C

Solution steps

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

Expand

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

= \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

= \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}

= ln ∣ y ∣ + \frac{y ^{2}}{2}

Add a constant to the solution

= ln ∣ y ∣ + \frac{y ^{2}}{2} + C

x^{2} + a x + b

\frac{d}{d x} (4 cos (π x))

\frac{693 x ^{\frac{5}{2}}}{8}

(x, y) \to (0, 0) lim (x y)

\frac{\partial}{\partial x} (e^{- (x - v t)^{2}})

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