Value of integral $\int \frac{(1-x)^{n+1}}{x} dx$

Question

I was working on a series, and I came up with integral, an indefinte form

$$\int \frac{(1-x)^{n+1}}{x} dx$$

I was wondering about how to solve this indefinte integral analytically. I solved it on Mathematica and got the result

so, I was thinking how to approach the problem and solve it.After seeing the answer,I got no clue about how to approach it.

Also, Mathematica always doesn't give the neat answer. Can anyone approach the problem step wise and give a more simple answer that a undergraduate can work with.

Answer 1

Please correct me if I've gone wrong but here goes.. $$ \int \frac{\left(1-x\right) \left(1-x\right)^n}{x}dx = \int \frac{\left(1-x\right)^{n+1}}{x}dx$$ $$ \int\frac{\left(1-x\right)^n}{x}dx - \int\left(1-x\right)^ndx$$ $$ \int\frac{\left(1-x\right)^n}{x}dx + \frac{\left(1-x\right)^{n+1}}{n+1} + k1$$ $$ lnx=t $$ $$dx = xdt$$ $$x= e^t$$ $$ \int\frac{\left(1-x\right)^n}{x}dx= \int(1- e^t)^ndt $$ $$\int(1- e^t)^ndt = \int[ 1 - {n \choose 1}e^t+ {n \choose 2}e^{2t}- {n \choose 3}e^{3t}+....+\left(-1\right)^n{n \choose n}e^{nt}]dt$$ $$= lnx- {n \choose 1}x+ \frac{{n \choose 2}x^2}{2}-\frac{ {n \choose 3}x^3}{3}+....+\frac{\left(-1\right)^n{n \choose n}x^n}{n}+k2$$

Hence, $$\int \frac{\left(1-x\right)^{n+1}}{x}dx= lnx- {n \choose 1}x+ \frac{{n \choose 2}x^2}{2}-\frac{ {n \choose 3}x^3}{3}+....+\frac{\left(-1\right)^n{n \choose n}x^n}{n}+ \frac{\left(1-x\right)^{n+1}}{n+1}+ k$$

Answer 2

My approach is similar to EtienneBfx's approach.

Let $u = (1 - x) \Rightarrow -du = dx.$

Also, note that
$[E_1] ~(u - 1) \times (u^n + u^{(n-1)} + \cdots + 1) = u^{(n+1)} - 1.$

Then, the (indefinite) integral becomes
$\int ~\frac{u^{(n+1)}}{1 - u} \times (-du) ~=~ \int \frac{u^{(n+1)}}{u - 1} \times du$
$= ~\int \frac{u^{(n+1)} - 1}{u - 1} \times du ~+~ \int \frac{1}{u - 1} \times du.$

In the line immediately above, the first term is easily resolved into a closed form expression via $E_1.$

Answer 3

The best way to do it, is to expand the formula \begin{equation} (1-x)^{n+1}=\sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^k x^k \end{equation} Then \begin{equation} \int \frac{(1-x)^{n+1}}{x} dx = \sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^k \int x^{k-1} dx \end{equation} Finnaly you can intergrate each term. \begin{equation} \int \frac{(1-x)^{n+1}}{x} dx = ln(x)-(n+1)x+\sum_{k=2}^{n+1} \binom{n+1}{k} (-1)^k \frac{x^{k}}{k-1} \end{equation}