What is the integral of (0.7y^3+10+2y^{-3}) ?

The integral of (0.7y^3+10+2y^{-3}) is 0.175y^4+10y-1/(y^2)+C

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integral of (0.7y^3+10+2y^{-3})

\int (0.7 y^{3} + 10 + 2 y^{- 3}) d y

0.175 y^{4} + 10 y - \frac{1}{y ^{2}} + C

Solution steps

\int 0.7 y^{3} + 10 + 2 y^{- 3} d y

2 y^{- 3} = \frac{2}{y ^{3}}

= \int 0.7 y^{3} + 10 + \frac{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 0.7 y^{3} d y + \int 10 d y + \int \frac{2}{y ^{3}} d y

\int 0.7 y^{3} d y = 0.175 y^{4}

\int 10 d y = 10 y

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

= 0.175 y^{4} + 10 y - \frac{1}{y ^{2}}

Add a constant to the solution

= 0.175 y^{4} + 10 y - \frac{1}{y ^{2}} + C

f (x) = x^{2} - 2 x - 3

\frac{d}{d x} (sin (cot (x)))

y = e^{x} cos (x)

\frac{7 s + 5}{( s + 2 ) ( s - 1 )}

\int_{0}^{1} (x^{3}) d x

What is the integral of (0.7y^3+10+2y^{-3}) ?
The integral of (0.7y^3+10+2y^{-3}) is 0.175y^4+10y-1/(y^2)+C