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

1 Answer
Feb 25, 2017

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