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

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

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integral of y-1/(y^2)

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

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

Solution steps

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

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

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

= \frac{y ^{2}}{2} - (- \frac{1}{y})

Simplify

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

Add a constant to the solution

= \frac{y ^{2}}{2} + \frac{1}{y} + C

2 t^{4}

\frac{d}{d x} (ln ((x e^{- 2 x})^{\frac{1}{3}}))

y = g (3 x) h (2 x)

\int x^{2} + a^{2} d x

x \to 0 lim (\frac{2 ( 5 x + 1 )}{5})