0
$\begingroup$

Does anyone know how to integrate this without using parts or u sub, just manipulating it as an indefinite integral. Like does it lok like you can use revrse product, chain, qoutient rule? Also all its x to the power of (-1) everywhere, sorry i didn't know how to put -1. I know ln is involved, but I can;t seem to figure out how to rearange things. Thanks! $$ \int \dfrac{x^{-1}}{1-x^{-1}}dx $$ EDIT: sorry I have the answer apparently I asked it a year ago!

$\endgroup$
1
$\begingroup$

you do realize that $x^{-1}=\frac{1}{x}$, right?

$$\int{\frac{\frac{1}{x}}{1-\frac{1}{x}}dx}=\int{\frac{1}{x-1}dx}=\ln(x-1)$$

$\endgroup$
  • $\begingroup$ Yeah, I wasn't thinking straight. But on a side note generally in math when we see a fraction and we see nothing obvious and there are two identical terms on the numerator and denominator should we always multiply by 1/1, x/x in this case. $\endgroup$ – mushimaster Mar 7 '18 at 17:35
  • $\begingroup$ You simply always try to simplify your expression in the most possible way, and if you notice a familiar integral, you evaluate it, if you don't then you proceed in the normal way (by parts or substitution or any other way), here i'm not multiplying by $'\frac{x}{x}'$ because it's helpful, or because there are two identical terms, i'm simplifying the expression more. $\endgroup$ – Mario SOUPER Mar 7 '18 at 17:39
  • $\begingroup$ Frankly, i'm just trying to get the 50 rep to be able to comment. $\endgroup$ – Mario SOUPER Mar 7 '18 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.