Normally what you do is select a
that will allow you to make the integral into something that you can actually integrate. In some cases we get given an integral with
terms on the top and the bottom of the fraction, so we normally select a
that when differentiated, will get rid of one of these terms (e.g.
, let
so that the derivative gets rid of the
expression, and gives us something that we are able to integrate).
Now in this case both of the expressions we have have the same largest power of
, so we need to choose the one that gives us a simpler expression when you rearrange it for
and substitute it in. In this case, we don't want to complicate the bottom, rater we want to make the denominator simpler, so that we can expand the fraction out later, so that is why we let
equal the bottom of the denominator.
e.g. If we tried to let
equal the numerator, we'd run into problems.
^{2}}dx<br />\\ \text{Let }u=2x-1<br />\\ \frac{du}{dx}=2<br />\\ \frac{1}{2}du=dx<br />\\ x=\frac{u+1}{2}<br />\\ \int\frac{2x-1}{\left(x-1\right)^{2}}dx & =\frac{1}{2}\int\frac{u}{\left(\frac{u+1}{2}\right)^{2}}du<br />\\ & =2\int\frac{u}{\left(u+1\right)^{2}}du<br />\end{alignedat}<br /> )
But if we let
equal the denominator:
^{2}}dx<br />\\ \text{Let }u=x-1<br />\\ \frac{du}{dx}=1<br />\\ du=dx<br />\\ x=u+1<br />\\ \int\frac{2x-1}{\left(x-1\right)^{2}}dx & =\int\frac{2\left(u+1\right)-1}{u^{2}}du<br />\\ & =\int\frac{2u+1}{u^{2}}du<br />\\ & =\int\frac{2}{u}+\frac{1}{u^{2}}du<br />\\ & =2\log_{e}|u|-\frac{1}{u}+C<br />\\ & =2\log_{e}|x-1|-\frac{1}{x-1}+C<br />\end{alignedat}<br /> )
Lazyred's method is probably the easiest.
If you wanted to do it by linear subst, you could do something like this
^2} \ dx = 2\int \frac{x}{(x-1)^2} \ dx - \int \frac{1}{(x-1)^2} \ dx)
and then make the substitution 
and note that 
and 
More or less what e^1 said.
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