How do you integrate by substitution #int (1+1/t)^3(1/t^2)dt#?

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Answer 1 · 2016-12-24T07:55:32

The answer is #=-1/4(1+1/t)^4+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C#

and #(x^n)'=nx^(n-1)#

We perform this integral by substitution

Let #u=1+1/t#, #=>#, #du = - 1 / ( t^2 ) dt#

Therefore,

#int(1+1/t)^3(1/t^2)dt#

#=intu^3*-du#

#=-u^4/4#

#=-1/4(1+1/t)^4+C#