Don't forget the +C

Trevor was wrong, I do in fact know how to evaluate the integral. Start by substituting $e^{2u}=v, 2e^{2u}du=dv, e^{2(\ln{N})}=N^2, e^{2(0)}=1$ .

The integral then becomes $\int \int_1^{N^2} \! \frac{2v-2}{9} \ dv \ dx = x+C$ .

Taking the derivative of both sides yields $\int_1^{N^2} \! \frac{2v-2}{9} \ dv = 1$ .

$\frac{v^2-2v}{9}|_1^{N^2} =1$

$\frac{N^4-2N^2+1}{9} =1$

$(N^2-1)^2=9$

Solving this gives that the only solution in the domain of $\ln{x}$ is $N=2$ .

This is a simple matter of noting that we wish $\displaystyle \frac{4}{9}\int\limits_{[0,\, \ln(N)]}(e^{4u}-e^{2u})~\mathrm{d}u=\frac{4}{9}\left(\frac{1}{4}(N^4-1)-\frac{1}{2}(N^2-1)\right)=\frac{1}{9}(N^4-2N^2-8)=1$ . Solve the quadratic for the positive value of $N^2=4$ , and then take the principal square root to get $N=\boxed{2}$ .