What is the integral of e^{2x}e^{3x} ?

The integral of e^{2x}e^{3x} is 1/5 e^{5x}+C

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integral of e^{2x}e^{3x}

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

\frac{1}{5} e^{5 x} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{\frac{5 u}{3}}}{3} d u

= \int \frac{e ^{\frac{5 u}{3}}}{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^{\frac{5 u}{3}} d u

Apply u-substitution

: \int e^{v} \frac{3}{5} d v

= \frac{1}{3} \cdot \int e^{v} \frac{3}{5} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{3}{5} \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{3} \cdot \frac{3}{5} e^{v}

Substitute back

= \frac{1}{3} \cdot \frac{3}{5} e^{\frac{5 \cdot 3 x}{3}}

Simplify

\frac{1}{3} \cdot \frac{3}{5} e^{\frac{5 \cdot 3 x}{3}} : \frac{1}{5} e^{5 x}

= \frac{1}{5} e^{5 x}

Add a constant to the solution

= \frac{1}{5} e^{5 x} + C

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

\int - t d t

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

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What is the integral of e^{2x}e^{3x} ?
The integral of e^{2x}e^{3x} is 1/5 e^{5x}+C