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

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

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integral of (y^2+1)/y

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

\frac{y ^{2}}{2} + ln ∣ y ∣ + C

Solution steps

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

Expand

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

= \int y + \frac{1}{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 y d y + \int \frac{1}{y} d y

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

\int \frac{1}{y} d y = ln ∣ y ∣

= \frac{y ^{2}}{2} + ln ∣ y ∣

Add a constant to the solution

= \frac{y ^{2}}{2} + ln ∣ y ∣ + C

\frac{d}{d x} ((x^{2} - 3 x + 4)^{4})

\frac{1}{1 - e ^{x}}

n = 0 \sum \infty \frac{3 + 2 n}{2 ^{n}}

n = 0 \sum \infty \frac{1}{n + 13}

\int \frac{y}{( y ^{2} - 4 ) ^{\frac{2}{5}}} d y

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