Question

How do you find the indefinite integral of `int (4/(3t^2)+7/(2t))dt`?

Answer 1

`int(4/(3t^2)+7/(2t))dt=-4/(3t)+7/2lnt+c`

Explanation:

As `intt^ndt=t^(n+1)/(n+1)+c`, but for `n=-1`, `intt^(-1)dt=int(dt)/t=lnt`

`int(4/(3t^2)+7/(2t))dt`

= `int(4/3t^(-2)+7/2t^(-1))dt`

= `4/3t^(-1)/(-1)+7/2lnt+c`

= `-4/(3t)+7/2lnt+c`