# Find $\int x^{-n}(1-x)^{-1}\,dx.$

Let $$f(x) = \frac{1}{x^n(1-x)}$$ where $n\in \mathbb{N}$ and $x$ is not 0 or 1.

In a previous question, i determined that the partial fraction decomposition of $f(x)$ is

$$f(x) = \sum_{j=1}^{n}\frac{1}{x^j} + \frac{1}{(1-x)}$$

Integrating this, when $n = 1$, I have:

$$\int f(x)\, dx = \ln|x|-\ln|1-x| + C = \ln\left|\frac{x}{1-x}\right|+C,$$ for $x \neq 1,0$.

If $n >1$, I have (after some trial and error),

$$\int f(x)\, dx = \ln|x|+ \sum_{j=2}^{n}\frac{x^{-j+1}}{-(j-1)}-\ln|1-x| + C$$

Have I missed anything here? Thanks for the feedback!

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Sure, it's good. I guess I don't like negatives in denominator, so would prefer $-\sum_{j=2}^n \frac{x^{-j+1}}{j-1}$. –  André Nicolas Oct 2 '12 at 23:16
+1 For a clear question with self work shown –  DonAntonio Oct 3 '12 at 3:12

For each $j\neq 1$, if

$$g_j(x)=x^{-j}$$

then

$$\int g_j(x) dx =\frac{x^{1-j}}{1-j}+C$$

Thus, given

$$f(x) = \sum_{j=1}^{n}\frac{1}{x^j} + \frac{1}{(1-x)}$$

we have

$$\int f(x)dx = \int\frac1 xdx+ \sum_{j=2}^{n}\int \frac{dx}{x^j} +\int \frac{dx}{(1-x)}$$

whence

$$\int f(x)dx = \log x+ \sum_{j=2}^{n}\frac{x^{1-j}}{1-j} +\log{(1-x)}+C$$

You're right.

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As a question of curiosity, when can we say $$\int \sum_{i}f_i(x) dx=\sum_i \int f_i(x)dx?$$ Is it always the case for indefinite integration? I recall having some information on this, but I've forgotten for some reason. –  000 Oct 2 '12 at 23:30
Disregard that; I found it: en.wikipedia.org/wiki/Fubini's_theorem –  000 Oct 2 '12 at 23:38
If the sum is finite, then you have no problem. For infinite sums, refer to this –  Pedro Tamaroff Oct 2 '12 at 23:39
@Limitless How does that relate to your issue? –  Pedro Tamaroff Oct 2 '12 at 23:40
I cannot articulate and portray a sincere understanding of the theorem (I've yet to cover all the concepts) in its entirety, but this answer seems to explain that: math.stackexchange.com/a/83747/22144 The intuitive understanding I received from that essentially comes down to this: If the two different expressions are not divergent, then the swapping of the summation and integral sign is justified. I'll look more carefully at your answer (that you just linked) in just a moment. . . –  000 Oct 4 '12 at 0:27