Đặt \(u=2^x\Rightarrow x=log_2u\left(=\dfrac{lnu}{ln2}\right)\Rightarrow dx=\dfrac{1}{u.ln2}.du\)
\(\Rightarrow I=\)\(\int\limits^2_1\dfrac{lnu}{ln2}.\dfrac{1}{u.ln2}.u.du=\)\(\dfrac{1}{ln^2}\int\limits^2_1lnu.du\)
Đặt \(\left\{{}\begin{matrix}m=lnu\\dn=du\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}dm=\dfrac{1}{u}.du\\n=u\end{matrix}\right.\)
=> \(I=\dfrac{1}{ln^22}\left[u.lnu|^2_1-\int\limits^2_1u.\dfrac{1}{u}du\right]=\dfrac{1}{ln^22}\left[2.ln2-1\right]=\dfrac{2.ln2-1}{ln^22}\)
\(S=2-1+1=2\)
Ý A
Đúng 1
Bình luận (0)
