What is the integral of (6x^2-4e^x) ?

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

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integral of (6x^2-4e^x)

\int (6 x^{2} - 4 e^{x}) d x

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

Solution steps

\int 6 x^{2} - 4 e^{x} 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 4 e^{x} d x

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

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

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

Add a constant to the solution

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

\frac{5 x ^{3} + 5 x ^{2}}{x}

\int \frac{arctan ( x )}{x ^{2}} d x

x = 4 y^{2}, x + y = 3

\int \frac{10 x}{81 x ^{4} + 49} d x

\int arccos (x) d x

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