What is the integral of (sqrt(y)-y)/(y^2) ?

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

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integral of (sqrt(y)-y)/(y^2)

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

- \frac{2}{y} - ln ∣ y ∣ + C

Solution steps

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

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

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

Simplify

\frac{y}{y ^{2}} - \frac{y}{y ^{2}} : \frac{1}{y ^{\frac{3}{2}}} - \frac{1}{y}

= \int \frac{1}{y ^{\frac{3}{2}}} - \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 \frac{1}{y ^{\frac{3}{2}}} d y - \int \frac{1}{y} d y

\int \frac{1}{y ^{\frac{3}{2}}} d y = - \frac{2}{y}

\int \frac{1}{y} d y = ln ∣ y ∣

= - \frac{2}{y} - ln ∣ y ∣

Add a constant to the solution

= - \frac{2}{y} - ln ∣ y ∣ + C

\int (ln (x))^{2} d x

\int_{- 1}^{1} ((1) - (x^{2})) d x

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

n = 2 \sum \infty \frac{1}{n 4 ^{n}}

6 e^{x} cos (x)

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