What is the integral of (e^{2x}+x) ?

The integral of (e^{2x}+x) is 1/2 e^{2x}+(x^2)/2+C

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integral of (e^{2x}+x)

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

\frac{1}{2} e^{2 x} + \frac{x ^{2}}{2} + C

Solution steps

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

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

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

= \frac{1}{2} e^{2 x} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{1}{2} e^{2 x} + \frac{x ^{2}}{2} + C

- 0.0000272 x^{2} + 0.008459 x + 440

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\frac{\partial}{\partial x} (sec (x))

\int 2 x (x^{2} + 12)^{3} d x

What is the integral of (e^{2x}+x) ?
The integral of (e^{2x}+x) is 1/2 e^{2x}+(x^2)/2+C