Question

What is the integral of 5(6t+e^(-t))(2t+4) with respect to t?

`int(5(6t+e^-t)(2t+4))dt`

Answer 1

`5int(6t+e^-t)(2t+4)dt=20t^3+60t^2-10te^-t-30e^-t+C`

Explanation:

Factor out `5` and multiply through.

`5int(6t+e^-t)(2t+4)dt`

`5int(12t^2+24t+2te^-t+4e^-t)dt`

We can split this sum up and integrate each term individually.

`5(int12t^2dt+int24tdt+int2te^-tdt+int4e^-tdt)`

`int12t^2dt=12/3t^3=4t^3`

`int24tdt=24/2t^2=12t^2`

`int4e^-tdt=-4e^-t`

So, the more complicated integral is `int2te^-tdt.` We'll have to integrate by parts.

`u=2t`

`du=2dt`

`dv=e^-tdt`

`v=-e^-t`

`uv-intvdu=-2te^-t+2inte^-tdt=-2te^-t-2e^-t`

Combining all of the above:

`5int(6t+e^-t)(2t+4)dt=5(4t^3+12t^2-2te^-t-2e^-t-4e^-t)+C`

Don't forget the constant of integration. Simplifying yields

`5int(6t+e^-t)(2t+4)dt=20t^3+60t^2-10te^-t-30e^-t+C`