What is the integral of (6x^2-8x+2) ?

The integral of (6x^2-8x+2) is 2x^3-4x^2+2x+C

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integral of (6x^2-8x+2)

\int (6 x^{2} - 8 x + 2) d x

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

Solution steps

\int 6 x^{2} - 8 x + 2 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 6 x^{2} d x - \int 8 x d x + \int 2 d x

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

\int 8 x d x = 4 x^{2}

\int 2 d x = 2 x

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

Add a constant to the solution

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

y^{^{'''}} + y^{^{'}} = cot (x)

y^{^{''}} + 5 \frac{d y}{d x} + 6 y = 2 e^{x}

n = 1 \sum \infty \frac{9 n}{2 n + 1}

f (x) = \frac{1}{arctan ( x ^{2} )}

5 arccot (t) + 5 arccot (\frac{1}{t})

What is the integral of (6x^2-8x+2) ?
The integral of (6x^2-8x+2) is 2x^3-4x^2+2x+C