Problem 4

If we define $F(a) \; = \; \int_0^1 \frac{x^a(x^4-1)}{\ln x}\,dx \;, \qquad \qquad a > - 1 \;,$ then $F'(a) \; = \; \int_0^1 x^a(x^4 - 1)\,dx \; = \; \frac{1}{a+5} - \frac{1}{a+1} \;, \qquad \qquad a > -1\;,$ and hence $F(a) \; = \; c + \ln(a+5) - \ln(a+1) \; =\; c + \ln\left(\frac{a+5}{a+1}\right) \;, \qquad \qquad a > -1 \;.$ The function $\frac{x^4-1}{\ln x}$ is integrable over $(0,1)$ and so, using the Dominated Convergence Theorem, it is clear that $\lim_{a\to\infty}F(a) \; = \; 0 \;,$ and hence $c=0$ , so that $F(a) \; = \; \ln\left(\frac{a+5}{a+1}\right) \;.$ We are asked to evaluate the integral $F(0) = \ln5$ , so the answer is $\boxed{5}$ .

U can do this in ur head..so easy