What is the integral of (4x^3+6x^2-1) ?

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

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integral of (4x^3+6x^2-1)

\int (4 x^{3} + 6 x^{2} - 1) d x

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

Solution steps

\int 4 x^{3} + 6 x^{2} - 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 4 x^{3} d x + \int 6 x^{2} d x - \int 1 d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int csc (\frac{x}{2}) d x

\frac{\partial}{\partial y} (10 x y)

\frac{1}{x ^{2} - 2 x - 3}

x \to 4 lim (\frac{1}{x + 4})

\int sin^{11} (x) cos^{5} (x) d x

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