Integral in terms of a parametric variable

Solution 1:

This is one of the most commonly used examples of differentiation under the integral sign or Feynman's Trick. Here, differentiate with respect to $a$ and that immediately gives you an easy-to-integrate expression.

The motive behind that is because when we differentiate $f(z)=z^a$ with respect to $a$, what's left is a natural log: $f'(z)=z^a\log z$. The $\log z$ term can then cancel with our denominator, leaving something trivial to deal with. To wit, differentiating gives$$I'(a)=\frac {\partial}{\partial a}\int\limits_0^1dx\,\frac {x^a-1}{\log x}=\int\limits_0^1dx\, x^a=\frac 1{a+1}$$Integrating the expression back to retrieve $I(a)$ leaves$$I(a)=\log(a+1)+C$$Now all we have to do is find the constant $C$. This can be easily found by finding a suitable value of $a$ such that $I(a)$ and $\log(a+1)$ can be easily evaluated. I will leave this up to the OP if he is interested in finishing this problem. Here is the final answer, hidden in all of its glory

$$\int\limits_0^1dx\,\frac {x^a-1}{\log x}\color{blue}{=\log(a+1)}$$

Here's an extra practice problem (slightly more difficult, but still feasible)$$I=\int\limits_0^{\infty}dx\,\frac {\log\left(\frac {1+x^{11}}{1+x^3}\right)}{(1+x^2)\log x}$$Anyone who solves it gets to feel good about themselves!