What is the integral of e^{3x}*sqrt(e^{3x)-4} ?

The integral of e^{3x}*sqrt(e^{3x)-4} is 2/9 (e^{3x}-4)^{3/2}+C

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integral of e^{3x}*sqrt(e^{3x)-4}

\int e^{3 x} \cdot e^{3 x} - 4 d x

\frac{2}{9} (e^{3 x} - 4)^{\frac{3}{2}} + C

Solution steps

\int e^{3 x} e^{3 x} - 4 d x

Apply u-substitution

: \int e^{u} e^{u} - 4 \frac{1}{3} d u

= \int e^{u} e^{u} - 4 \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} e^{u} - 4 d u

Apply u-substitution

: \int v d v

= \frac{1}{3} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= \frac{1}{3} \cdot \frac{2}{3} (e^{3 x} - 4)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (e^{3 x} - 4)^{\frac{3}{2}} : \frac{2}{9} (e^{3 x} - 4)^{\frac{3}{2}}

= \frac{2}{9} (e^{3 x} - 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (e^{3 x} - 4)^{\frac{3}{2}} + C

t e^{- t} cos (5 t)

f (x) = sin (9 x) + cos (2 x)

\frac{d}{d x} (sec (x) - tan (x))

x \to 0 lim (sech^{2} (x))

\frac{\partial}{\partial r} (r sin (t))

What is the integral of e^{3x}*sqrt(e^{3x)-4} ?
The integral of e^{3x}*sqrt(e^{3x)-4} is 2/9 (e^{3x}-4)^{3/2}+C