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

1 Answer

Shwetank Mauria

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#

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