Integrating : $\int_0^1 {\frac {x^a-x^b} {\ln x} dx}$ We are given parameters $a > 0, b > 0$.
Task is to integrate that:
$\displaystyle \int_0^1 {\frac {x^a-x^b} {\ln x} dx}$.
I have tried approaching problem from different angles with no luck. 
I tried integration by parts(tried all combinations of possible $v$ and $u$), u-substitution with no luck.
Also I tried to integrate this two similar terms separately.
Tried to get some idea of how to go from answer, got nice answer from MATLAB: $\displaystyle \ln{\frac{a+1}{b+1}}$, but no idea how to reach it.
I would appreciate some suggestions.
 A: First notice that:
$$I = \int_{0}^{1} \frac{x^b-x^a}{\ln x} dx =  \int_{0}^{1} \Big[\int_{a}^{b} x^y dy\Big]dx $$
The function $f(x,y)=x^y$ is continous in the set $[0,1]\times[a,b]$, therefore:
$$ I= \int_{0}^{1} \Big[\int_{a}^{b} x^y dy\Big]dx = \int_{a}^{b} \Big[\int_{0}^{1} x^y dx\Big]dy = \int_{a}^{b} \frac{1}{y+1}dy = \ln \Big(\frac{b+1}{a+1} \Big)$$
A: If we let $t=-\ln x$ then, $dx =-e^{-t}dt$ and the result follows frullani's theorem
$$ \int_{0}^{1} \frac{x^b-x^a}{\ln x} dx =- \int_{0}^{\infty} \frac{e^{-tb}-e^{-ta}}{t} e^tdt=-\int_{0}^{\infty} \frac{e^{-t(b+1)}-e^{-t(a+1)}}{t} dt= \ln\left(\frac{b+1}{a+1}\right)$$
Here is the Frullani's theorem :Proof of Frullani's theorem
Also see this  https://math.stackexchange.com/q/2521520
A: First of all I got to say that I like user140835's solution much more. It has this strict pure beauty of mathematical derivations.
But, from one hand, I have a feeling that I would fail to recognize applicability of that approach on similar problems. From other hand, I wanted to solve it by myself with  another approach suggested in comments.
So I will leave my solution here, just in case if it will be helpful for somebody.  
Performing u-substitution: $u = \frac 1 x$. Hence $dx = \frac {-du} {u^2}$
$\displaystyle -\int \frac {u^{-a}-u^{-b}}{u^2 ln{\frac 1 u}}du = I\space \Rightarrow \space \frac {dI(a)} {da} = \frac {u^{-a} \cdot ln(u)} {u^2ln{\frac 1 u}}$ 
Now, integrate received expression over u: $\int_\infty^0 \frac{dI(a)}{da}du=\int_\infty^0 \frac {u^{-a} \cdot ln(u)} {u^2ln{\frac 1 u}}du=-\frac {u^{-(a+1)}} {a+1} |_\infty^0=\frac {-1} {a+1}$.
Now we have to find antiderivative of that, which is $I(a)= -ln(a+1)$.
Following analogous steps for $I(b)$ we get the answer: $log(\frac{b+1}{a+1})$
