Why is approach 2 wrong here? I am relatively newe to calculus please explain in detail what is going wrong... Evaluate: $ F(\alpha) = \int_0^1 \frac{x^\alpha-1}{\ln x} dx $
Approach 1: (Correct)
Using Newton Leibniz Theorem, \begin{align} F'(\alpha) &= \int_0^1 \frac{\delta}{\delta\alpha} \left(\frac{x^\alpha-1}{\ln x}\right) dx \\ \implies F'(\alpha) &= \int_0^1 x^{\alpha} dx\\ \implies F'(\alpha) &= \frac{1}{\alpha+1} \\ \implies F(\alpha) &= \ln (\alpha+1) + c \end{align}
We can get $ c=0 $ from $ F(0)=0 $, Hence $$ F(\alpha) = \ln (\alpha+1)$$
Approach 2: (Incorrect) $$F(\alpha) = \int_0^1 \frac{x^\alpha}{\ln x} dx - \int_0^1 \frac{1}{\ln x} dx $$ Substituite $ x^{\alpha+1}=t $ $$ F(\alpha) = \int_0^1 \frac{1}{\ln t} dt - \int_0^1 \frac{1}{\ln x} dx $$ $$ F(\alpha) = 0 $$