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Find the integral of (2x-1)/(x-1)^2 dx.
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I suppose we have to split it into the sum of two integrals?
How?
\[\int\frac{2x}{(x-1)^2}dx-\int\frac1{(x-1)^2}dx\]
And then?
u=x-1
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Integration by substitution
I mean, du=dx?
Yes
Wait don't split now
Split after substituting
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\[\begin{array}{cl} &\int\frac{2x-1}{(x-1)^2}dx\\ =&\int\frac{2u+1}{u^2}du\\ =&\int\frac2udu+\int\frac1{u^2}du \end{array}\]
I suppose you can do it now?
I'll try.
Remember the power law (or however you call it): \(\Large\displaystyle\int x^ndx=\frac{x^{n+1}}{n+1}+C\)
I got it. Thanks.
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no problem
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