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

\frac{d}{d x} ((x^{3} - x + 1)^{10})

y = \frac{1}{x ^{4}}, (1, 1)

\int \frac{x ^{6}}{( 3 - x ^{7} ) ^{2}} d x

\int \frac{- 6}{( x ^{2} + 6 ) ^{\frac{3}{2}}} d x

x \to 0 + lim ((1 + \frac{2}{x})^{x})

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