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How do you integrate #int (lnx)^5/x# using substitution?

Answer 1

Christopher Allers

#= (ln x)^6/6 + C #

for starters, there's a clear pattern here:

#d/dx ( ln^n x) = n ln^(n-1) x 1/x#

#int (lnx)^5/x \ dx = int d/dx ( 1/6ln^6 x) \ dx = 1/6ln^6 x + C#

but if you are required to introduce substitution into this, which seems OTT, I suppose you could say that

#u = ln x, du = 1/x dx #

so you have

#int u^5/x \ x \ du #

#= int u^5 \ du #

#= u^6/6 + C #

#= (ln x)^6/6 + C #

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Answer 2

Aurora Alvarado

To integrate ( \int \frac{{(\ln(x))^5}}{x} , dx ) using substitution, let ( u = \ln(x) ). Then, ( du = \frac{1}{x} dx ). The integral becomes ( \int u^5 , du ). Integrating ( u^5 ) with respect to ( u ) yields ( \frac{1}{6} u^6 + C ), where ( C ) is the constant of integration. Substituting back ( u = \ln(x) ) gives ( \frac{1}{6} (\ln(x))^6 + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.