Proving $\int_0^1\frac{dx}{[ax+b(1-x)]^2}=\frac1{ab}$ 
Let $a,b$ be real numbers.$$I=\displaystyle\int_0^1\frac1{[ax+b(1-x)]^2}dx=\frac1{ab}$$

Before starting to calculate this integral, we need to investigate consistence of it. For example, it is obvious that if we take $a,b$ so that making $ab<0$ we get from positive integral to negative result (because $\frac1{[ax+b(1-x)]^2}$ is always positive). 
Attempt:
I thought that the only singularity is $x=\frac{b}{b-a}$, but the ratio $\frac{b}{b-a}$ can be bigger than $1$, if we take $a=1,b=3$ and can be less than $0$ if we take $a=3,b=1$ so we have 2 things.
First, if $0<\frac{b}{b-a}<1$ we must consider the integral as improper one, second, if the ratio $\frac{b}{b-a}\in(-\infty,0)\cup(1,\infty)$ we do not have to think because we take integral over $[0,1]$
If we can calculate first type, the second type will be equal to it, so let us calculate.

Note that if $n\neq -1$: $$\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+C$$
  Hence: $$\displaystyle\int\frac1{[ax+b(1-x)]^2}dx=\frac{((a-b)x+b)^{-1}}{-a}+C$$

So;
$$I=\lim\limits_{u\to\left(\frac{b}{b-a}\right)^-}\left\{\frac{((a-b)x+b)^{-1}}{-a}\right\}_0^u+\lim\limits_{u\to\left(\frac{b}{b-a}\right)^+}\left\{\frac{((a-b)x+b)^{-1}}{-a}\right\}^1_u$$
I thought this was not good method so, I did :
$$I=\lim\limits_{\epsilon\to 0}\left\{\frac{((a-b)x+b)^{-1}}{-a}\right\}_0^{\left(\frac{b}{b-a}\right)-\epsilon}+\lim\limits_{\epsilon\to0}\left\{\frac{((a-b)x+b)^{-1}}{-a}\right\}^1_{\left(\frac{b}{b-a}\right)+\epsilon}$$
so;
$$I=\left(\frac1{ab}-\frac1{a^2}\right)+\lim\limits_{\epsilon\to 0}\left(\frac2{\epsilon a(a-b)}\right)$$

But this is not what I wanted to find, so what am I supposed to do and follow, what is my problem? Thank you in advanced.

 A: Simpler version just See that :
*If $ab>0$ then there is no singularity and we have 
$$\int_0^1\frac{dx}{[ax+b(1-x)]^2}\overset{\color{blue}{t=(a-b)x+b}}{=}\frac{1}{b-a}\int_a^b\frac{dt}{t^2} =\color{red}{\frac1{ab}} $$
*If $ab\le 0$ then there is a singularity at $x=\frac{b}{b-a}$ ot at $t=0$ in this case the integral does not exists and  its principal value is given by 
$$pv\int_0^1\frac{dx}{[ax+b(1-x)]^2}\overset{\color{blue}{t=(a-b)x+b}}{=}\frac{pv}{b-a}\int_a^b\frac{dt}{t^2} \\=\lim_{\varepsilon\to 0}\frac{1}{b-a}\left(\int_a^{-\varepsilon}\frac{dt}{t^2}+\int_{\varepsilon}^b\frac{dt}{t^2}\right)\\=\lim_{\varepsilon\to 0}\frac{1}{b-a} \left(\frac{1}{a}+\frac{1}{\varepsilon} -\frac{1}{b}+\frac{1}{\varepsilon}\right) =\color{red}{\infty} $$
In fact in this last case either $a\le0\le b$ or  $b\le0\le a$ with $a\neq b$

A: Interesting Historical Note : This integral is useful for loop calculations in Quantum Electrodynamics. My understanding is that its introduction to the subject was due to Richard Feynman, and I have heard the integral referred to as "Feynman's Trick". 
What follows is the calculation for a fraction with a general number of factors in the denominator. It is assumed that all of these factors are positive. 
Consider the fraction $\frac{1}{D_1 D_2\cdots D_n }$ with $D_i>0$ for each $i=1,2,\dots,n$. 
Note that $\int_0^\infty dt\ e^{-tD} = \frac{1}{D}$. We can then write, 
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^\infty \cdots \int_0^\infty  dt_1 \cdots dt_n e^{-\sum t_i D_i }  $$
Introduce a variable $T\geq 0 $ which represents the sum of the $t_i$. This can be accomplished by introducing $1=\int_0^\infty dT\ \delta(T-\sum t_i)$ 
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^\infty \cdots \int_0^\infty  dt_1 \cdots dt_n \int_0^\infty dT \delta\left(T-\sum t_i\right)e^{-\sum t_i D_i }  $$
Now introduce $x_i=t_i/T$, where $\sum x_i = 1$. 
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^\infty \cdots \int_0^\infty  dt_1 \cdots dt_n \int_0^\infty dT \delta\left(T\cdot (1-\sum x_i )\right)e^{-T \sum x_i D_i }  $$
Use the identity for delta functions $\delta(ax)=\frac{1}{a} \delta(x)$ for $a>0$. 
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^\infty \cdots \int_0^\infty  dt_1 \cdots dt_n \int_0^\infty dT \frac{1}{T}\delta\left(1-\sum x_i \right)e^{-T \sum x_i D_i }  $$
Now change variables from $t_i$ to $x_i$. 
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^1 \cdots \int_0^1  dx_1 \cdots dx_n \int_0^\infty dT \frac{T^n}{T}\delta\left(1-\sum x_i \right)e^{-T \sum x_i D_i }  $$
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^1 \cdots \int_0^1  dx_1 \cdots dx_n \int_0^\infty dT \ T^{n-1}\delta\left(1-\sum x_i \right)e^{-T \sum x_i D_i }  $$
Rescale the integral over $dT$ by introducing a new variable $\xi=T \sum x_i D_i$ . 
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^1 \cdots \int_0^1  dx_1 \cdots dx_n \int_0^\infty d\xi \frac{\xi^{n-1}}{\left(\sum x_i D_i\right)^n}\delta\left(1-\sum x_i \right)e^{-\xi  }  $$
$$\frac{1}{D_1 D_2\cdots D_n } = \int_0^1 \cdots \int_0^1 dx_1 \cdots dx_n  \frac{1}{\left(\sum x_i D_i \right)^n} \delta\left(1-\sum x_i \right) \int_0^\infty d\xi \  \xi^{n-1}e^{-\xi  }  $$
The integral over $\xi$ is just the gamma function.
$$\frac{1}{D_1 D_2\cdots D_n } = \Gamma(n) \int_0^1 \cdots \int_0^1  dx_1 \cdots dx_n  \frac{1}{\left(\sum x_i D_i \right)^n} \delta\left(1-\sum x_i \right)  $$
We can use the $\delta$ function to perform the $x_n$ integration, this has the effect of replacing $x_n$ with $1-\sum_{i=1}^n x_i$. 
$$\boxed{\frac{1}{D_1 D_2\cdots D_n } = \Gamma(n) \int_0^1 \cdots \int_0^1  dx_1 \cdots dx_{n-1}  \frac{1}{\left(D_n(1-\sum_{i=1}^{n-1} x_i) + \sum_{i=1}^{n-1} x_i D_i \right)^n}   }$$
