What is the integral of (0.3y^4-8y^{-3}+2) ?

The integral of (0.3y^4-8y^{-3}+2) is 0.06y^5+4/(y^2)+2y+C

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integral of (0.3y^4-8y^{-3}+2)

\int (0.3 y^{4} - 8 y^{- 3} + 2) d y

0.06 y^{5} + \frac{4}{y ^{2}} + 2 y + C

Solution steps

\int 0.3 y^{4} - 8 y^{- 3} + 2 d y

8 y^{- 3} = \frac{8}{y ^{3}}

= \int 0.3 y^{4} - \frac{8}{y ^{3}} + 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 0.3 y^{4} d y - \int \frac{8}{y ^{3}} d y + \int 2 d y

\int 0.3 y^{4} d y = 0.06 y^{5}

\int \frac{8}{y ^{3}} d y = - \frac{4}{y ^{2}}

\int 2 d y = 2 y

= 0.06 y^{5} - (- \frac{4}{y ^{2}}) + 2 y

Simplify

= 0.06 y^{5} + \frac{4}{y ^{2}} + 2 y

Add a constant to the solution

= 0.06 y^{5} + \frac{4}{y ^{2}} + 2 y + C

x \to - 1 + lim (x)

\frac{d}{d x} (\frac{( 1 + x ^{2} + x ^{4} )}{1 + x + x ^{2}})

y^{^{'}} - \frac{9}{x} y = \frac{y ^{5}}{x ^{20}}

y^{^{''}} + 16 y = sin (4 t)

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

What is the integral of (0.3y^4-8y^{-3}+2) ?
The integral of (0.3y^4-8y^{-3}+2) is 0.06y^5+4/(y^2)+2y+C