Solving an Integral Problem: Finding the Minus Sign

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SUMMARY

This discussion focuses on understanding the presence of a negative sign in the integral of the function involving the natural logarithm. The user initially believes the integral should yield ln^2(x-1)/2, but through the application of the chain rule and u-substitution (u = 1 - x), it is clarified that the correct result is -ln^2(x-1)/2. The negative sign arises from the derivative of the logarithmic function, confirming the importance of recognizing the impact of function transformations in integral calculus.

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Homework Statement



2e2gumw.png

(ignore what's written, it isn't important for the problem)

I'm studying integrals and I came across this solved example. However I can't understand where the minus of the integral came from came from.

Homework Equations





The Attempt at a Solution



The primitive of a f^p * f ' function is f^(p+1) / (p+1) Therefore it should be ln^2(x-1)/2 and not -ln^2(x-1)/2

Thanks
 
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You could dispel any doubts by making a u-substitution. Try [itex]u = 1 - x[/itex] and see where that takes you.
 
Jalo said:

Homework Statement



2e2gumw.png

(ignore what's written, it isn't important for the problem)

I'm studying integrals and I came across this solved example. However I can't understand where the minus of the integral came from came from.

Homework Equations



The Attempt at a Solution



The primitive of a f^p * f ' function is f^(p+1) / (p+1) Therefore it should be ln^2(x-1)/2 and not -ln^2(x-1)/2

Thanks

By the chain rule: [itex]\displaystyle \frac{d}{dx}\ln(1-x)=\frac{1}{1-x}\frac{d}{dx}(1-x)=-\frac{1}{1-x}\,.[/itex]

So you have the anti-derivative of f^p * (- f ' ) .
 

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