`int t e^(-3t) dt` Evaluate the integral

You need to use the substitution -`3t = u` , such that:

`-3t=  u => -3dt = du => dt = -(du)/3`

Replacing the variable, yields:

`int t*e^(-3t) dt = (1/9)int u*e^u du`

You need to use the integration by parts such that:

`int fdg = fg - int gdf`

`f = u => df = du`

`dg = e^u=> g = e^u`

`(1/9)int u*e^u du = (1/9)(u*e^u - int e^u du)`

`(1/9)int u*e^u du = (1/9)(u*e^u - e^u) + c`

Replacing back the variable, yields:

`int t*e^(-3t) dt = (1/9)((-3t)*e^(-3t) - e^(-3t)) + c`

Hence, evaluating the integral, using substitution, then integration by parts, yields `int t*e^(-3t) dt = ((e^(-3t))/9)(-3t - 1) + c`