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Simple Integration

  1. Dec 22, 2013 #1
    Could anyone please tell me why is the integration of ∫(x+2)/(x-1)dx ≠ (x-1)+3ln(x-1).

    I got it using substitution of u=x+1.
  2. jcsd
  3. Dec 22, 2013 #2


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    I'd do the integral as follows
    [tex]\int \mathrm{d} x \frac{x+2}{x-1}=\int \mathrm{d} x \frac{x-1+3}{x-1} = \int \mathrm{d} x \left (1+\frac{3}{x-1} \right )=x+3 \ln(|x-1|)+C.[/tex]
    So up to your missing modulus under the log (which only means that your result is valis for [itex]x>1[/itex] only), you got the correct solution. Why do you think it's wrong?
  4. Dec 22, 2013 #3
    I was confused.

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