What is the integral of (9z^2+14z-2) ?

The integral of (9z^2+14z-2) is 3z^3+7z^2-2z+C

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integral of (9z^2+14z-2)

\int (9 z^{2} + 14 z - 2) d z

3 z^{3} + 7 z^{2} - 2 z + C

Solution steps

\int 9 z^{2} + 14 z - 2 d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 z^{2} d z + \int 14 z d z - \int 2 d z

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

\int 14 z d z = 7 z^{2}

\int 2 d z = 2 z

= 3 z^{3} + 7 z^{2} - 2 z

Add a constant to the solution

= 3 z^{3} + 7 z^{2} - 2 z + C

\int \frac{3 x ^{2}}{x ^{2} + 4 x + 5} d x

\int \frac{1}{x ^{2}} \cdot (\frac{1}{t} - 1)^{\frac{1}{2}}

\int \frac{1}{( sin ( a ) y ) ( cos ( a ) y )} d y

\int (\frac{1}{3 x + 2}) d x

\int \frac{x ^{2} - 1}{x ^{3} - 5 x ^{2} + x - 5} d x

What is the integral of (9z^2+14z-2) ?
The integral of (9z^2+14z-2) is 3z^3+7z^2-2z+C