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

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

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

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

\frac{1}{3} e^{3 x + 2} + C

Solution steps

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

Simplify

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

= \int e^{3 x + 2} d x

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= \int e^{u} \frac{1}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{3} e^{u}

Substitute back

u = 3 x + 2

= \frac{1}{3} e^{3 x + 2}

Add a constant to the solution

= \frac{1}{3} e^{3 x + 2} + C

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