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

I am try to integrate the following function:$$\frac{1}{x(1-x)^n}$$ I have searched on some online integral calculator, but they are not solving it. Can we use binomial theorem in this?

• Have you tried partial fractions? Commented Oct 20, 2019 at 9:19
• The antiderivative can always be given (for integer $n$) in terms of logs and rational functions via partial fraction decomposition. If you are dead-set on using the Binomial Theorem, note $\frac{1}{x}(1-x)^{-n} = \frac{1}{x}\sum_{k=0}^\infty {n+k-1 \choose k}x^k$ so that $\int\frac{1}{x}(1-x)^{-n} = \sum_{k=0}^\infty {n+k-1 \choose k}\int x^{k-1} = \sum_{k=0}^\infty {n+k-1 \choose k}k^{-1}x^k$ Commented Oct 20, 2019 at 9:20

Not sure about the Binomial theorem, but if you make substitution $$y=1-x$$, it becomes $$-\int \frac{1}{(1-y)y^n}$$. Now notice that $$\frac{1}{(1-y)y^n} = \frac{1}{y^n}+\frac{1}{y^{n-1}}+\dots+\frac{1}{y}+\frac{1}{1-y}$$ (you might want to think why this is). So you have finitely many terms you should be able to integrate.

To include non-integer $$n$$, Maple uses a hypergeometric $$\int \frac{dx}{x(1-x)^n} = nx\;\mbox{_3F_2}\big(1,1,1+n;\,2,2;\,x\big)+\ln \left( x \right) + C$$

Assuming $$n\in\mathbb{N}_+$$, let

$$I=\int\frac{1}{x\left(1-x\right)^{n}}\mathop{\mathrm{d}x}$$

Do partial fraction decomposition

$$\frac{1}{x\left(1-x\right)^{n}}=\frac{A_{0}}{x}+\frac{A_{1}}{1-x}+\frac{A_{2}}{\left(1-x\right)^{2}}+\dots+\frac{A_{n}}{\left(1-x\right)^{n}}$$

Finding out $$A_{0}$$ is easy. Multiply both sides of equation with $$x$$

$$\frac{1}{\left(1-x\right)^{n}}=A_{0}+\frac{A_{1}x}{1-x}+\frac{A_{2}x}{\left(1-x\right)^{2}}+\dots+\frac{A_{n}x}{\left(1-x\right)^{n}}$$

Let $$x=0$$ and you can get $$A_{0}=1$$.

To find out the value of the rest coefficients, move $$\frac{A_{0}}{x}$$ to the left side and multiply both sides with $$(1-x)^{n}$$

$$\frac{1}{x}-\frac{\left(1-x\right)^{n}}{x}=A_{1}\left(1-x\right)^{n-1}+A_{2}\left(1-x\right)^{n-2}+\dots+A_{n-1}\left(1-x\right)+A_{n}$$

Let $$x=1$$ and you can get $$A_{n}=1$$.

To find out $$A_{n-1}$$, move $$A_{n}$$ to the left side and divide both sides by $$1-x$$

$$\frac{\frac{1}{x}-1-\frac{\left(1-x\right)^{n}}{x}}{1-x}=\frac{1}{x}-\frac{\left(1-x\right)^{n-1}}{x}=A_{1}\left(1-x\right)^{n-2}+A_{2}\left(1-x\right)^{n-3}+\dots+A_{n-2}\left(1-x\right)+A_{n-1}$$

Let $$x=1$$ and you can get $$A_{n-1}=1$$.

Similarly you can get $$A_{n-2}=A_{n-3}=\dots=A_{2}=A_{1}=1$$.

Now calculate the integral

\begin{aligned} I&=\int\left(\frac{1}{x}+\frac{1}{1-x}+\frac{1}{\left(1-x\right)^{2}}+\dots+\frac{1}{\left(1-x\right)^{n}}\right)\mathop{\mathrm{d}x}\\ &=\int\frac{1}{x}\mathop{\mathrm{d}x}+\int\frac{1}{1-x}\mathop{\mathrm{d}x}+\sum_{k=2}^{n}\int\frac{1}{\left(1-x\right)^{k}}\mathop{\mathrm{d}x}\\ &=\boxed{\ln\left(\left|x\right|\right)-\ln\left(\left|1-x\right|\right)+\sum_{k=2}^{n}\frac{1}{\left(k-1\right)\left(1-x\right)^{k-1}}+C} \end{aligned}