Question

How do you integrate `int (1+3t)t^2dt`?

Answer 1

The answer is `=t^3/3+(3t^4)/4+C`

Explanation:

We use, `intx^ndx=x^(n+1)/(n+1)+C (x!=-1)`

So,

`int(1+3t)t^2dt=int (t^2+3t^3)dt`

`=intt^2dt+3intt^3dt`

`=t^3/3+3t^4/4+C`