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

The integral of (3^x-x^3+e^x) is (3^x}{ln(3)}-\frac{x^4)/4+e^x+C

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integral of (3^x-x^3+e^x)

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

\frac{3 ^{x}}{ln ( 3 )} - \frac{x ^{4}}{4} + e^{x} + C

Solution steps

\int 3^{x} - x^{3} + 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 3^{x} d x - \int x^{3} d x + \int e^{x} d x

\int 3^{x} d x = \frac{3 ^{x}}{ln ( 3 )}

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

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

= \frac{3 ^{x}}{ln ( 3 )} - \frac{x ^{4}}{4} + e^{x}

Add a constant to the solution

= \frac{3 ^{x}}{ln ( 3 )} - \frac{x ^{4}}{4} + e^{x} + C

x \to 0 lim (- 3 x^{2} - 5 x + 1)

\int \frac{7 - x ^{2}}{5 x ^{3} + x} d x

x \to 0 lim (x e^{- \frac{1}{x}})

f (x) = 8 x, at x = 16

lo g_{3} (2 x^{2} + 4 x + 3)

What is the integral of (3^x-x^3+e^x) ?
The integral of (3^x-x^3+e^x) is (3^x}{ln(3)}-\frac{x^4)/4+e^x+C