Deceptive Integral

We have $I \; = \; \int_0^1 \frac{(2^x+1)^2 - 2 \times 2^x}{2^x+1} \; = \; \int_0^1 \left(2^x + 1 - 2 \frac{2^x}{2^x+1}\right)\,dx \; = \; \frac{1}{\ln2} + 1 - 2\int_0^1 \frac{1}{2^x+1}\,2^x\,dx$ Using the substitution $y = 2^x$ , we obtain $I \; = \; \frac{1}{\ln2} + 1 - \frac{2}{\ln2}\int_1^2\frac{1}{y+1}\,dy \; = \; \frac{1}{\ln2} + 1 - \frac{2}{\ln2}\Big[\ln(y+1)\Big]_1^2 \; = \; \frac{1 - \ln\frac98}{\ln2} \; = \; \boxed{1.272770039}$