What is the integral of (16x^3-14x+6) ?

The integral of (16x^3-14x+6) is 4x^4-7x^2+6x+C

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integral of (16x^3-14x+6)

\int (16 x^{3} - 14 x + 6) d x

4 x^{4} - 7 x^{2} + 6 x + C

Solution steps

\int 16 x^{3} - 14 x + 6 d x

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 x^{3} d x - \int 14 x d x + \int 6 d x

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

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

\int 6 d x = 6 x

= 4 x^{4} - 7 x^{2} + 6 x

Add a constant to the solution

= 4 x^{4} - 7 x^{2} + 6 x + C

\int 6^{x} x d x

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\int \frac{sec ^{2} ( x )}{1 + 9 tan ^{2} ( x )} d x

What is the integral of (16x^3-14x+6) ?
The integral of (16x^3-14x+6) is 4x^4-7x^2+6x+C