What is the integral of (3sqrt(y)-2y^{-3}) ?

The integral of (3sqrt(y)-2y^{-3}) is 2y^{3/2}+1/(y^2)+C

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integral of (3sqrt(y)-2y^{-3})

\int (3 y - 2 y^{- 3}) d y

2 y^{\frac{3}{2}} + \frac{1}{y ^{2}} + C

Solution steps

\int 3 y - 2 y^{- 3} 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 3 y d y - \int 2 y^{- 3} d y

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

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

= 2 y^{\frac{3}{2}} - (- \frac{1}{y ^{2}})

Simplify

= 2 y^{\frac{3}{2}} + \frac{1}{y ^{2}}

Add a constant to the solution

= 2 y^{\frac{3}{2}} + \frac{1}{y ^{2}} + C

x \to \infty lim (\frac{ln ( 3 x + 5 )}{ln ( 7 x + 3 ) + 1})

\frac{d x}{d t} + 7 x = 5 cos (2 t)

n = 1 \sum \infty \frac{n ^{3}}{4 ^{n}}

x \to 1 lim ([g (x) - 3 f (x)])

\int e^{y} d y

What is the integral of (3sqrt(y)-2y^{-3}) ?
The integral of (3sqrt(y)-2y^{-3}) is 2y^{3/2}+1/(y^2)+C