What is the integral of 2te^{2t}+1/3 e^t ?

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

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integral of 2te^{2t}+1/3 e^t

\int 2 t e^{2 t} + \frac{1}{3} e^{t} d t

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

Solution steps

\int 2 t e^{2 t} + \frac{1}{3} 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 2 t e^{2 t} d t + \int \frac{1}{3} e^{t} d t

\int 2 t e^{2 t} d t = e^{2 t} t - \frac{1}{2} e^{2 t}

\int \frac{1}{3} e^{t} d t = \frac{1}{3} e^{t}

= e^{2 t} t - \frac{1}{2} e^{2 t} + \frac{1}{3} e^{t}

Add a constant to the solution

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

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