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

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

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integral of (e^{2x}+1)

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

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

Solution steps

\int e^{2 x} + 1 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 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

\int \frac{2 x + 4}{x ^{2} + 4 x + 8} d x

\int x + 2 d x

x d x + sin^{2} (x^{- 1} y) (y d x - x d y) = 0

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

tan (2 x - x^{3})

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