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Mathematics 14 Online
OpenStudy (anonymous):

Find the integral of (2x-1)/(x-1)^2 dx.

OpenStudy (kc_kennylau):

I suppose we have to split it into the sum of two integrals?

OpenStudy (anonymous):

How?

OpenStudy (kc_kennylau):

\[\int\frac{2x}{(x-1)^2}dx-\int\frac1{(x-1)^2}dx\]

OpenStudy (anonymous):

And then?

OpenStudy (kc_kennylau):

u=x-1

OpenStudy (kc_kennylau):

Integration by substitution

OpenStudy (anonymous):

I mean, du=dx?

OpenStudy (kc_kennylau):

Yes

OpenStudy (kc_kennylau):

Wait don't split now

OpenStudy (kc_kennylau):

Split after substituting

OpenStudy (kc_kennylau):

\[\begin{array}{cl} &\int\frac{2x-1}{(x-1)^2}dx\\ =&\int\frac{2u+1}{u^2}du\\ =&\int\frac2udu+\int\frac1{u^2}du \end{array}\]

OpenStudy (kc_kennylau):

I suppose you can do it now?

OpenStudy (anonymous):

I'll try.

OpenStudy (kc_kennylau):

Remember the power law (or however you call it): \(\Large\displaystyle\int x^ndx=\frac{x^{n+1}}{n+1}+C\)

OpenStudy (anonymous):

I got it. Thanks.

OpenStudy (kc_kennylau):

no problem

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