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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{1 + u ^{2}}{u} d u

Expand

\frac{1 + u ^{2}}{u} : \frac{1}{u} + u

= \int \frac{1}{u} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u} d u + \int u d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

= ln ∣ u ∣ + \frac{u ^{2}}{2}

Substitute back

u = e^{x}

= ln ∣ e^{x} ∣ + \frac{( e ^{x} ) ^{2}}{2}

Simplify

ln ∣ e^{x} ∣ + \frac{( e ^{x} ) ^{2}}{2} : x + \frac{1}{2} e^{2 x}

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

Add a constant to the solution

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

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