What is the integral of 16y^3+9y^2-6y+3 ?

The integral of 16y^3+9y^2-6y+3 is 4y^4+3y^3-3y^2+3y+C

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integral of 16y^3+9y^2-6y+3

\int 16 y^{3} + 9 y^{2} - 6 y + 3 d y

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

Solution steps

\int 16 y^{3} + 9 y^{2} - 6 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 16 y^{3} d y + \int 9 y^{2} d y - \int 6 y d y + \int 3 d y

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

\int 9 y^{2} d y = 3 y^{3}

\int 6 y d y = 3 y^{2}

\int 3 d y = 3 y

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

Add a constant to the solution

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

\int_{0}^{1} 4^{4 x} d x

\int \frac{cos ( x )}{sin ^{2} ( x ) - 64} d x

\int cos (2 π) d x

\frac{d}{d x} (\frac{cos ( 6 x )}{e ^{x}})

What is the integral of 16y^3+9y^2-6y+3 ?
The integral of 16y^3+9y^2-6y+3 is 4y^4+3y^3-3y^2+3y+C