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

The integral of e^{3x}sqrt(1+2e^{3x)} is 1/9 (1+2e^{3x})^{3/2}+C

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

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

\frac{1}{9} (1 + 2 e^{3 x})^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{6} d u

= \int \frac{u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 2 e^{3 x}

= \frac{1}{6} \cdot \frac{2}{3} (1 + 2 e^{3 x})^{\frac{3}{2}}

Simplify

\frac{1}{6} \cdot \frac{2}{3} (1 + 2 e^{3 x})^{\frac{3}{2}} : \frac{1}{9} (1 + 2 e^{3 x})^{\frac{3}{2}}

= \frac{1}{9} (1 + 2 e^{3 x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9} (1 + 2 e^{3 x})^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (1 - 2 x^{2} - 2 y)

(1 - t^{2}) y^{^{''}} - 2 t y^{^{'}} + 2 y = 0

\frac{d}{d x} ((x + a) e^{2 x})

\int_{0}^{4} x e^{x} d x

\frac{d}{d t} (\frac{t ^{3}}{6} + \frac{t ^{2}}{4} + \frac{1}{2} t)

What is the integral of e^{3x}sqrt(1+2e^{3x)} ?
The integral of e^{3x}sqrt(1+2e^{3x)} is 1/9 (1+2e^{3x})^{3/2}+C