# How to integrate $\int_0^x \frac{1+\epsilon X}{1-X}~dX$?

I'm having a hard time trying to figure out the steps to get to the final answer shown below. $$\int_0^x\frac{1+\epsilon X}{1-X}dX = (1+\epsilon)\ln\frac{1}{1-X}-\epsilon X$$

Any help would be great.

I attached what I've been getting...

• I tried using u substitution but I keep getting epsilon(1-X)-(epsilon+1)*ln(1-x).... Dec 13, 2014 at 3:04
• I think I can guess what happened here, but it would be clearer if you posted your $u$-substitution as part of the question. Dec 13, 2014 at 4:02
• Thanks David. I added a picture of what I've been getting...Could you help me figure out what I'm doing wrong? I used u substitution...The other answers are not really helping me figure this out... Dec 13, 2014 at 4:36
• why is the integral from (1-x) to 1? I understand (1-x) but not the 1. Thanks! Dec 13, 2014 at 4:58
• @Integrator ohh got it! that was what I missed. thank you so much! I've spent way too much time on this 1 problem. Dec 13, 2014 at 5:28

Split into two integrals:

$$\int_0^x \frac{1}{1-x}dx \;+\; \epsilon\int_0^x \frac{ x}{1-x}dx$$ The rest is basic integration and then applying the fundamental theorem of calculs part II.

• This is the straightforward answer...I don't understand why one needs complicate this with substitutions... Dec 13, 2014 at 3:28
• To do the integrals you will have to do a substitution. Dec 13, 2014 at 3:33
• One could use long division for the second integral to simplify the expression $\frac{x}{1-x}=\frac{1}{1-x}-1$ Dec 13, 2014 at 3:37

You can let $y = 1 - X$, the denominator. So $X = 1 - y$ and $dx = - dy$. The integral becomes $$-\int_1^{1 - x} \frac{1 + \epsilon(1 - y)}{y}\,dy$$ $$-(1 + \epsilon)\int_1^{1-x} {1 \over y}\,dy + \epsilon \int_1^{1-x} 1 \,dy$$ I think you can take it from here.

Replace $1-X$ by $Y$, then your integral becomes $\displaystyle \int_0^x\frac{1+\epsilon X}{1-X}dX\, = \int_{1-x}^{1}\frac{1+\epsilon (1-Y)}{Y} dY\,$ . Now it is easy to integrate

\begin{align*} \int\limits_0^x {\frac{{1 + \varepsilon X}} {{1 - X}}dX} &= \int\limits_0^x {\frac{{1 + \varepsilon - \varepsilon \left( {1 - X} \right)}} {{1 - X}}dX} = \left( {1 + \varepsilon } \right)\int\limits_0^x {\frac{1} {{1 - X}}dX} - \varepsilon \int\limits_0^x {dX}\\ &= - \left( {\varepsilon + 1} \right)\int\limits_0^x {\frac{1} {{X - 1}}dX} - \varepsilon x = - \left( {\varepsilon + 1} \right)\left. {\left( {\ln \left| {X - 1} \right|} \right)} \right|_{X = 0}^{X = x} - \varepsilon x\\ &= - \left( {\varepsilon + 1} \right)\ln \left|\left( {x - 1} \right)\right| - \varepsilon x = \left( {\varepsilon + 1} \right)\ln \left( {\left| {\frac{1} {{x - 1}}} \right|} \right) - \varepsilon x. \end{align*}

• Thanks for the explanation! Could you explain how you got from step 1 to step 2? Dec 13, 2014 at 3:19
• This is just technique. $1 + \varepsilon - \varepsilon \left( {1 - X} \right) = 1 + \varepsilon - \varepsilon + \varepsilon X = 1 + \varepsilon X$. Dec 13, 2014 at 3:27