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I previously asked a different question about the same problem. In my textbook, the steps to solving $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ are as follows:

$$\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$$ $$\int \frac{2x^2-4x+3}{x^2-2x+1} \, dx$$ $$\int 2+\frac{1}{(x-1)^2}\, dx$$ $$\int 2dx+\frac{dx}{(x-1)^2} $$

$$\int 2x-\frac{1}{(x-1)}+C$$

I don't understand how for step 4 there can be two dx's for one integral.

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    $\begingroup$ $\int [f (x)+g(x)]\, dx =\int f(x)\, dx+\int g(x)\, dx$. What exactly is your confusion? $\endgroup$ Commented Jul 23, 2019 at 6:20
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    $\begingroup$ If your textbook literally wrote that, then get a better textbook! $\endgroup$ Commented Jul 23, 2019 at 6:23

3 Answers 3

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The two final steps should be$$\int2\,\mathrm dx+\int\frac{\mathrm dx}{(x-1)^2}$$and$$2x-\frac1{x-1}+C$$respectively.

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It's $$\int2dx+\int\frac{1}{(x-1)^2}dx$$

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Your notebook seems to be making a mistake, the $dx$ should have been written after the integral is split into two parts

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