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 \frac{( x + 1 ) ^{3}}{x} d x

\int \frac{7}{9} sec^{2} (\frac{x}{9}) d x

\int (4 cosh (x) + x^{5}) d x

\int x^{2} + z^{2} d z

\int \frac{x ^{2}}{( x ^{3} + 9 ) ^{2}} 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