What is the integral of e^{sqrt(3x+9)} ?

The integral of e^{sqrt(3x+9)} is 2/3 (e^{sqrt(3x+9)}sqrt(3x+9)-e^{sqrt(3x+9)})+C

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integral of e^{sqrt(3x+9)}

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

\frac{2}{3} (e^{3 x + 9} 3 x + 9 - e^{3 x + 9}) + C

Solution steps

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

Apply u-substitution

= \int \frac{2 e ^{u} u}{3} d u

Take the constant out:

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

= \frac{2}{3} \cdot \int e^{u} u d u

Apply Integration By Parts

= \frac{2}{3} (e^{u} u - \int e^{u} d u)

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

= \frac{2}{3} (e^{u} u - e^{u})

Substitute back

u = 3 x + 9

= \frac{2}{3} (e^{3 x + 9} 3 x + 9 - e^{3 x + 9})

Add a constant to the solution

= \frac{2}{3} (e^{3 x + 9} 3 x + 9 - e^{3 x + 9}) + C

What is the integral of e^{sqrt(3x+9)} ?
The integral of e^{sqrt(3x+9)} is 2/3 (e^{sqrt(3x+9)}sqrt(3x+9)-e^{sqrt(3x+9)})+C