I need to test the convergence of the integral $$\int_{0}^{1} \frac{x^n}{1-x} dx$$
Now, $f(x) = \frac{x^n}{1-x}$ and it has point of infinite discontinuity at $x = 1$ . If we do the $\mu$ test, then, $$\lim_{x\to1} (1-x)^{1} \frac{x^n}{1-x} = 1$$ which is finite and non-zero. Here, $\mu = 1$ but the condition of convergence of the integral is $0<\mu<1$ , and here, since $\mu = 1$ , the integral diverges. I do not have a solution and this is all that I derived. Is my approach correct? If not, then what is?
Your solution is correct. Let me describe some intuition: As $x\to 1^-$, the numerator of the integrand tends to $1$ while the denominator tends to zero. Since the denominator is $1-x$ and $x\to 1^-$, the fraction blows up at the same rate that $\frac{1}{u}$ does as $u\to 0^+$. So we should use the divergence of $\int_0^1 \frac{1}{u}\,du$ to decide the divergence of $\int_0^1 \frac{x^n}{1-x}\,dx$.
Now, to be more formal: Make the substitution $u=1-x$. This transforms $\int_0^1 \frac{x^n}{1-x}\,dx$ to $\int_0^1 \frac{(1-u)^n}{u}\,du$. Using the binomial theorem, the integrand simplifies to a polynomial in $u$, plus $\frac{1}{u}$. There is no question that the polynomial part is integrable on $0 \leq u \leq 1$. So the convergence of $\int_0^1 \frac{(1-u)^n}{u}\,du$ is equivalent to the convergence of $\int_0^1 \frac{1}{u}\,du$, and we know that this integral diverges. Thus $\int_0^1 \frac{(1-u)^n}{u}\,du$ diverges, meaning that $\int_0^1 \frac{x^n}{1-x}\,dx$ diverges.