What is the integral of (16y^3-3y^2+2) ?

The integral of (16y^3-3y^2+2) is 4y^4-y^3+2y+C

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integral of (16y^3-3y^2+2)

\int (16 y^{3} - 3 y^{2} + 2) d y

4 y^{4} - y^{3} + 2 y + C

Solution steps

\int 16 y^{3} - 3 y^{2} + 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 16 y^{3} d y - \int 3 y^{2} d y + \int 2 d y

\int 16 y^{3} d y = 4 y^{4}

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

\int 2 d y = 2 y

= 4 y^{4} - y^{3} + 2 y

Add a constant to the solution

= 4 y^{4} - y^{3} + 2 y + C

\int \frac{x ^{4} - x ^{3}}{4 x ^{5} - 5 x ^{4}} d x

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

\int \frac{- 4 x}{2 + 2 x ^{2}} d x

\int \frac{csc ( y )}{1} d y

lo g_{10} (\frac{1}{x})

What is the integral of (16y^3-3y^2+2) ?
The integral of (16y^3-3y^2+2) is 4y^4-y^3+2y+C