is the answer to this integral, 1/ (1-x)
= ln l 1 - x l
or
= - ln l 1 - x l
if its - ln l1-x l , could you explain why.
thanks.
Hello, johntuan!
Is the answer to this integral: .$\displaystyle \int\frac{dx}{1-x}$
. . $\displaystyle \ln|1 - x| + C\;\text{ or }\;-\ln |1-x| + C$ ?
If it is $\displaystyle -\ln |1-x|+C$, could you explain why?
We have: .$\displaystyle \int\frac{dx}{1-x}$
Let: .$\displaystyle u \:=\:1-x\quad\Rightarrow\quad du \:=\:{\color{red}-}dx \quad\Rightarrow\quad dx \:=\:{\color{red}-}du$
Substitute: .$\displaystyle \int\frac{{\color{red}-}du}{u} \;=\;{\color{red}-}\int\frac{du}{u} \;=\;{\color{red}-}\ln|u| + C$
Back-substitute: .$\displaystyle {\color{red}-}\ln|1-x| + C$
Just in case a picture helps...
Straight continuous lines differentiate downwards (integrate up) with respect to x, the straight dashed line similarly but with respect to the dashed balloon expression, so that the triangular network, on the right, satisfies the chain rule.
Don't integrate - balloontegrate!
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