I have a challenging integral in front of me, which has evaded all types of substitutions and seems to be intractible to me(P.S.:I am a high schooler). Here it is:
$$\int_0^1 \frac{x^a-1}{\ln (x)}$$
My teacher says that it can be solved by the method of differentiation. Is he referring to differentiation under the integral sign (which I have only heard about)? How do I go about this question?
I have tried the substitutions $x^a=t$ and then $\ln (t)=u$ which has led to a more compact form.
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
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!