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Popular Calculus >

integral of 1/y+y

Solution

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

Solution

ln ∣ y ∣ + \frac{y ^{2}}{2} + C

Solution steps

\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

Graph

Popular Examples

integral of x^4)

\int d (x^{4}) d x

integral of (sqrt(-9x^2+4))/(x^2)

\int \frac{- 9 x ^{2} + 4}{x ^{2}} d x

integral of ((sqrt(x^2-16))/x)

\int (\frac{x ^{2} - 16}{x}) d x

integral of 2e^{8x}

\int 2 e^{8 x} d x

integral of (10x^3-5x)\sqrt[5]{x^4-x^2}

\int (10 x^{3} - 5 x) 5 x^{4} - x^{2} d x

Frequently Asked Questions (FAQ)

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