Graph Part of a line

Question

The formula and graph below get close to what I want. This integral shows the AREA of 0.1 to 0.7 in this formula, but I only want the line SEGMENT from 0.1 to 0.7 How do I do that?

Formula:$$ \int _{0.1}^{0.7}\frac{x}{1-x}dx$$

Integral of a formula graph

Answer 1

If you want to calculate the area between $$y=\frac{x}{1-x}$$ graph, the $x=0.1$ line, the $x=0.7$ line and the x axis:

Let $$P=\int_{0.1}^{0.7}\frac{x}{1-x}dx$$ Now substitute: $t=1-x \Rightarrow x=1-t \Rightarrow dt=\frac{dt}{dx}dx=-dx \Rightarrow dx=-dt$ $$P=\int_{0.1}^{0.7}\frac{1-t}{t}\cdot (-dt)=-\int_{0.1}^{0.7}\frac{1-t}{t}dt=-(\int_{0.1}^{0.7}\frac{1}{t}dt-\int_{0.1}^{0.7}dt)=[-(ln(t)-t)+C]_{0.1}^{0.7}=[t-ln(t)+C]_{0.1}^{0.7} $$ Now let's undo substitution: $$P = [1-x-ln(1-x)+C]_{0.1}^{0.7} =1-0.7-ln(0.3)-1+0.1+ln(0.9)=ln(0.9)-ln(0.3)-0.6 \approx 0.49861228866$$

If you want to calculate the curve length from $x=0.1$ to $x=0.7$ then you should compute $L=\int_{0.1}^{0.7}\sqrt{(dx)^2+(dy)^2}$ We know that $y=\frac{x}{1-x}$, therefore $dy=\frac{dy}{dx}dx=\frac{d \frac{x}{1-x}}{dx}dx=\frac{d \text{ }x \cdot (\frac{1}{x} \circ (1-x))}{dx}dx=(\frac{-1}{(1-x)^2}\cdot(-1)\cdot x + \frac{1}{1-x})dx=(\frac{x}{(1-x)^2}+\frac{1}{1-x})dx=\frac{x+1-x}{(1-x)^2}dx=\frac{1}{(1-x)^2}dx$ Therefore: $$L=\int_{0.1}^{0.7}\sqrt{(dx)^2+(\frac{1}{(1-x)^2}dx)^2}=\int_{0.1}^{0.7}\sqrt{1+\frac{1}{(1-x)^4}}dx \approx 2.332071129605535$$

Answer 2

To make life easier $$\frac x {1-x}=\frac{x-1+1}{1-x}=-1+\frac1{1-x}$$

$$\int_a^b\frac x {1-x}\,dx=-\int_a^b dx+\int_a^b\frac {dx}{1-x}= ???$$