What is the integral of (3t-5)e^{2t} ?

The integral of (3t-5)e^{2t} is 3/2 e^{2t}t-13/4 e^{2t}+C

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integral of (3t-5)e^{2t}

\int (3 t - 5) e^{2 t} d t

\frac{3}{2} e^{2 t} t - \frac{13}{4} e^{2 t} + C

Solution steps

\int (3 t - 5) e^{2 t} d t

Expand

(3 t - 5) e^{2 t} : 3 e^{2 t} t - 5 e^{2 t}

= \int 3 e^{2 t} t - 5 e^{2 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 3 e^{2 t} t d t - \int 5 e^{2 t} d t

\int 3 e^{2 t} t d t = \frac{3}{4} (2 e^{2 t} t - e^{2 t})

\int 5 e^{2 t} d t = \frac{5}{2} e^{2 t}

= \frac{3}{4} (2 e^{2 t} t - e^{2 t}) - \frac{5}{2} e^{2 t}

Simplify

\frac{3}{4} (2 e^{2 t} t - e^{2 t}) - \frac{5}{2} e^{2 t} : \frac{3}{2} e^{2 t} t - \frac{13}{4} e^{2 t}

= \frac{3}{2} e^{2 t} t - \frac{13}{4} e^{2 t}

Add a constant to the solution

= \frac{3}{2} e^{2 t} t - \frac{13}{4} e^{2 t} + C

What is the integral of (3t-5)e^{2t} ?
The integral of (3t-5)e^{2t} is 3/2 e^{2t}t-13/4 e^{2t}+C