What is the integral of (16/2 x^3+12x^5-2x+1) ?

The integral of (16/2 x^3+12x^5-2x+1) is 2x^4+2x^6-x^2+x+C

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integral of (16/2 x^3+12x^5-2x+1)

\int (\frac{16}{2} x^{3} + 12 x^{5} - 2 x + 1) d x

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

Solution steps

\int \frac{16}{2} x^{3} + 12 x^{5} - 2 x + 1 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 \frac{16}{2} x^{3} d x + \int 12 x^{5} d x - \int 2 x d x + \int 1 d x

\int \frac{16}{2} x^{3} d x = 2 x^{4}

\int 12 x^{5} d x = 2 x^{6}

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

\int 1 d x = x

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

Add a constant to the solution

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

f (x) = 5 e^{x} (cos (x) - sin (x))

\frac{\partial}{\partial x} ((x^{2} + y^{2})^{0.5})

y^{^{''}} = 1

f (2) = x^{3} - 2 x

x \to 2 - lim (x + 7)

What is the integral of (16/2 x^3+12x^5-2x+1) ?
The integral of (16/2 x^3+12x^5-2x+1) is 2x^4+2x^6-x^2+x+C