What is the integral of te^{3t}-e^t ?

The integral of te^{3t}-e^t is 1/9 (e^{3t}*3t-e^{3t})-e^t+C

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integral of te^{3t}-e^t

\int t e^{3 t} - e^{t} d t

\frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) - e^{t} + C

Solution steps

\int t e^{3 t} - e^{t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t e^{3 t} d t - \int e^{t} d t

\int t e^{3 t} d t = \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t})

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

= \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) - e^{t}

Add a constant to the solution

= \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) - e^{t} + C

\int \frac{1}{x ^{2} - 12} d x

\int \frac{1}{x 25 x ^{2} - 25} d x

x \to - 5.8 lim (ln (x^{2} - 9))

n = 1 \sum \infty \frac{n ^{2}}{n ^{3}}

\frac{1}{4} x

What is the integral of te^{3t}-e^t ?
The integral of te^{3t}-e^t is 1/9 (e^{3t}*3t-e^{3t})-e^t+C