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

The integral of (3e^x-3^x+3) is 3e^x-(3^x)/(ln(3))+3x+C

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

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

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

f (x) = - 5 x^{3} - 7 x^{2} + 8 x - 7

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

\frac{d}{d x} (x ln (x) + 2 x)

\frac{\partial}{\partial x} (x^{2} ln (x^{2} + y^{2}))

What is the integral of (3e^x-3^x+3) ?
The integral of (3e^x-3^x+3) is 3e^x-(3^x)/(ln(3))+3x+C