What is the integral of (e^{3x}+1)/(e^x+1) ?

The integral of (e^{3x}+1)/(e^x+1) is 1/2 e^{2x}-e^x+x+C

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integral of (e^{3x}+1)/(e^x+1)

\int \frac{e ^{3 x} + 1}{e ^{x} + 1} d x

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

Solution steps

\int \frac{e ^{3 x} + 1}{e ^{x} + 1} d x

Simplify

\frac{e ^{3 x} + 1}{e ^{x} + 1} : e^{2 x} - e^{x} + 1

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

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

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

\int 1 d x = x

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

Add a constant to the solution

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

x \to 3 lim (2 x^{3} - 4 x - 10)

9 x^{2} x + 7 x^{2}

(- 9, - 2), (3, - 1)

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

\frac{- 1}{s ( s - 2 )}

What is the integral of (e^{3x}+1)/(e^x+1) ?
The integral of (e^{3x}+1)/(e^x+1) is 1/2 e^{2x}-e^x+x+C