Integration: 𝛿(x+2)/(x-1)dx = (x-1)+3ln(x-1)

crack_head · Dec 22, 2013

Could anyone please tell me why is the integration of âˆ«(x+2)/(x-1)dx â‰ (x-1)+3ln(x-1).

I got it using substitution of u=x+1.

vanhees71 · Dec 22, 2013

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?

crack_head · Dec 22, 2013

I was confused.

Thanks

Integration: 𝛿(x+2)/(x-1)dx = (x-1)+3ln(x-1)

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Integration: 𝛿(x+2)/(x-1)dx = (x-1)+3ln(x-1)

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