We recognize this is the required formula:

`inte^(au)\ sin\ bu\ du` `=(e^(au)(a\ sin\ bu-b\ cos\ bu))/(a^2+b^2)+K`

For this example, we use:

`a = 2`

`b = 3`

`u = x`

So

`inte^(2x)sin\ 3x\ dx`

`=(e^(2x)(2\ sin 3x-3\ cos\ 3x))/(2^2+3^2)+K`

`=(e^(2x)(2\ sin\ 3x-3\ cos\ 3x))/13+K`

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