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

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

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integral of e^{3t}t+e^t

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

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

Solution steps

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

\int e^{3 t} 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

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