Biết I = intlimits_0^1 dx/2^x+1 = {log _a}b tính S=a+3b

Đặt $t = {2^x} \Rightarrow dt = {2^x}.\ln 2dx \Leftrightarrow {2^x}dx = \frac{{dt}}{{\ln 2}}.$ 

Đổi cận: $\left\{ \begin{array}{l} x = 0 \to t = 1\\ x = 1 \to t = 2 \end{array} \right.$  

Khi đó  $I = \int\limits_0^1 {\frac{{dx}}{{{2^x} + 1}}} = \int\limits_0^1 {\frac{{{2^x}dx}}{{{2^x}.({2^x} + 1)}} = \frac{1}{{\ln 2}}\int\limits_1^2 {\frac{{dt}}{{t(t + 1)}}} = \frac{1}{{\ln 2}}\int\limits_1^2 {\left( {\frac{1}{t} - \frac{1}{{t + 1}}} \right)dt = \frac{1}{{\ln 2}}.\ln \left. {\left| {\frac{t}{{t + 1}}} \right|} \right|_1^2} }$$\Rightarrow I = \frac{1}{{\ln 2}}.\left( {\ln \frac{2}{3} - \ln \frac{1}{2}} \right) = \frac{{\ln \frac{4}{3}}}{{\ln 2}} = {\log _2}\left( {\frac{4}{3}} \right)$

 mà $I = {\log _a}b \Rightarrow \left\{ \begin{array}{l} a = 2\\ b = \frac{4}{3} \end{array} \right. \Rightarrow S = a + 3b = 6.$