What is the integral of (2x^3+6e^x) ?

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

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integral of (2x^3+6e^x)

\int (2 x^{3} + 6 e^{x}) d x

\frac{x ^{4}}{2} + 6 e^{x} + C

Solution steps

\int 2 x^{3} + 6 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 2 x^{3} d x + \int 6 e^{x} d x

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

\int 6 e^{x} d x = 6 e^{x}

= \frac{x ^{4}}{2} + 6 e^{x}

Add a constant to the solution

= \frac{x ^{4}}{2} + 6 e^{x} + C

\int \frac{x ^{2} + 1}{( x + 1 ) ^{2}} d x

\int (- 3 x^{2} + 2 x + 5) d x

\frac{d}{d x} (2 x^{3} + 6 x y^{2} - 3 x^{2} + 3 y^{2})

cos (3 x), 2.5

1 - t

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