The one with the definite integral in $x$ !

Let's transform the given integral to the form of, $I(a)=\int_{0}^{1}\frac {x^a-1}{\ln x}$ with the given limits.The given integral equals $\ln (a+1)$ .For this we shall transform the inner term as $\frac {x^a-1}{\ln x}=\int_{y=0}^{a}x^ydy)$ .Now that we have two integrals we may use Fubinis Theorem to switch our integrals.The rest is easy and it yields the required result as $\ln (a+1)$ .Now set $a=7,I=\ln 8=3\ln 2=2.0794$ Aliter-Feynmans Trick