What is the integral of-e^{2t}+te^t ?

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

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integral of-e^{2t}+te^t

\int - e^{2 t} + t e^{t} d t

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

Solution steps

\int - e^{2 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^{2 t} d t + \int t e^{t} d t

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

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

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

Add a constant to the solution

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

\int_{0}^{1} x 6 - x^{2} d x

\frac{d}{d x} (sin (x) + \frac{3}{4} cot (x))

\frac{\partial}{\partial x} (cos (x y^{2}))

y^{^{''}} + 4 y = e^{(- t)}

csc (x + 5)

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