Nguyên hàm lớp 12 hàm số mũ có hướng dẫn giải

$I=\displaystyle\int\limits {{\dfrac{dx}{1+8^x}}}=\displaystyle\int\limits {{\dfrac{8^xdx}{8^x\left({1+8^x}\right)}}}$

Đặt $t=8^x$ ta có: $dt=8^x.\ln 8dx$ hay$8^xdx=\dfrac1{\ln 8}dt$

Ta có: $I=\dfrac1{\ln 8}\displaystyle\int\limits {{\dfrac{dt}{t\left({t+1}\right)}}}$$=\dfrac1{\ln 8}\displaystyle\int\limits {{\dfrac{\left({t+1}\right)-t}{t\left({t+1}\right)}dt}}$

$=\dfrac1{\ln 8}\displaystyle\int\limits {{\left({\dfrac1t-\dfrac1{t+1}}\right)dt}}$ $=\dfrac1{\ln 8}\left({\ln \left|t\right|-\ln \left|{t+1}\right|}\right)+C$

$=\dfrac1{\ln 8}.\ln \left|{\dfrac{t}{t+1}}\right|+C$ $=\dfrac1{\ln 8}\ln \left({\dfrac{8^x}{8^x+1}}\right)+C$