What is the integral of e^{3t}(e^{3t}-1) ?

The integral of e^{3t}(e^{3t}-1) is 1/6 (e^{3t}-1)^2+C

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integral of e^{3t}(e^{3t}-1)

\int e^{3 t} (e^{3 t} - 1) d t

\frac{1}{6} (e^{3 t} - 1)^{2} + C

Solution steps

\int e^{3 t} (e^{3 t} - 1) d t

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = e^{3 t} - 1

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

Simplify

\frac{1}{3} \cdot \frac{( e ^{3 t} - 1 ) ^{2}}{2} : \frac{1}{6} (e^{3 t} - 1)^{2}

= \frac{1}{6} (e^{3 t} - 1)^{2}

Add a constant to the solution

= \frac{1}{6} (e^{3 t} - 1)^{2} + C

What is the integral of e^{3t}(e^{3t}-1) ?
The integral of e^{3t}(e^{3t}-1) is 1/6 (e^{3t}-1)^2+C