What is the integral of (te^ti-e^{-2t}j+te^{t^2}k) ?

The integral of (te^ti-e^{-2t}j+te^{t^2}k) is 1/2 (je^{-2t}+ke^{t^2})+i(-e^t+e^tt)+C

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integral of (te^ti-e^{-2t}j+te^{t^2}k)

\int (t e^{t} i - e^{- 2 t} j + t e^{t^{2}} k) d t

\frac{1}{2} (j e^{- 2 t} + k e^{t^{2}}) + i (- e^{t} + e^{t} t) + C

Solution steps

\int t e^{t} i - e^{- 2 t} j + t e^{t^{2}} k 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^{t} i d t - \int e^{- 2 t} j d t + \int t e^{t^{2}} k d t

\int t e^{t} i d t = i (e^{t} t - e^{t})

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

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

= i (e^{t} t - e^{t}) - (- \frac{1}{2} j e^{- 2 t}) + k \frac{1}{2} e^{t^{2}}

Simplify

i (e^{t} t - e^{t}) - (- \frac{1}{2} j e^{- 2 t}) + k \frac{1}{2} e^{t^{2}} : \frac{1}{2} (j e^{- 2 t} + k e^{t^{2}}) + i (- e^{t} + e^{t} t)

= \frac{1}{2} (j e^{- 2 t} + k e^{t^{2}}) + i (- e^{t} + e^{t} t)

Add a constant to the solution

= \frac{1}{2} (j e^{- 2 t} + k e^{t^{2}}) + i (- e^{t} + e^{t} t) + C

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What is the integral of (te^ti-e^{-2t}j+te^{t^2}k) ?
The integral of (te^ti-e^{-2t}j+te^{t^2}k) is 1/2 (je^{-2t}+ke^{t^2})+i(-e^t+e^tt)+C