What is the integral of e^{x/3+9} ?

The integral of e^{x/3+9} is 3e^{9+x/3}+C

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integral of e^{x/3+9}

\int e^{\frac{x}{3} + 9} d x

3 e^{9 + \frac{x}{3}} + C

Solution steps

\int e^{\frac{x}{3} + 9} d x

e^{\frac{x}{3} + 9} = e^{9} e^{\frac{x}{3}}

= \int e^{9} e^{\frac{x}{3}} d x

Take the constant out:

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

= e^{9} \cdot \int e^{\frac{x}{3}} d x

Apply u-substitution

: \int e^{u} \cdot 3 d u

= e^{9} \cdot \int e^{u} \cdot 3 d u

Take the constant out:

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

= e^{9} \cdot 3 \cdot \int e^{u} d u

Use the common integral:

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

= e^{9} \cdot 3 e^{u}

Substitute back

u = \frac{x}{3}

= e^{9} \cdot 3 e^{\frac{x}{3}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{9} e^{\frac{x}{3}} = e^{9 + \frac{x}{3}}

= 3 e^{9 + \frac{x}{3}}

Add a constant to the solution

= 3 e^{9 + \frac{x}{3}} + C

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