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,

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

$=intu^3*-du$

$=-u^4/4$

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

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