Integration 44

Notice that $\displaystyle\int_3^7 x^t\,dt = \frac{x^7-x^3}{\ln(x)}$ so we have $\displaystyle\int_0^1 \frac{x^7-x^3}{\ln(x)} \,dx =\displaystyle\int_0^1\displaystyle\int_3^7x^t\,dt\,dx$ then using Fubini's theorem we interchange the order of integration to give $\displaystyle\int_3^7\displaystyle\int_0^1x^t\,dx\,dt$ which reduces to $\displaystyle\int_3^7 \frac{1}{t+1}\,dt$ which is easily evaluated as $\ln(t+1)\big|_{t=3}^{t=7} = \ln(8)-\ln(4) = \ln(2)$

$\begin{aligned}\text{Let } I_a&=\int_0^1\frac{x^{7a}-x^{3a}}{\ln x}dx\\ \therefore\frac{d}{da}I_a&=\int_0^1\frac{d}{da}\frac{x^{7a}-x^{3a}}{\ln x}dx\quad\text{by differentiation under the integral sign}\\ &=\int_0^1\frac{7x^{7a}\ln x-3x^{3a}\ln x}{\ln x}dx\\ &=\int_0^1\left(7x^{7a}-3x^{3a}\right)dx\\ &=\left[\frac{7x^{7a+1}}{7a+1}-\frac{3x^{3a+1}}{3a+1}\right]_0^1\\ &=\frac{7}{7a+1}-\frac{3}{3a+1}\\ \therefore I_a&=\int\left(\frac{7}{7a+1}-\frac{3}{3a+1}\right)da\\ &=\ln(7a+1)-\ln(3a+1)+C\\ &\text{Now, when }a=0,\,I_0=0,\,\text {as }\int_0^1\frac{x^0-x^0}{\ln x}=0\\ \text{So }0&=\ln(0+1)-\ln(0+1)+C\\ \therefore C&=0\\ \therefore I_a&=\ln(7a+1)-\ln(3a+1)\\ \therefore I_1&=\ln(7+1)-\ln(3+1)\\ \therefore\int_0^1\frac{x^7-x^3}{\ln x}dx&=\boxed{\ln(2)\approx 0.693}\end{aligned}$