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…

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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…

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.

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*}