Đặt \(I=\int\limits^{\ln5}_{\ln3}\frac{dx}{e^x+2e^{-x}-3}\) và \(t=e^x\). Trong các khẳng định sau, khẳng định nào sai?
\(I=\ln3-\ln2\) \(\dfrac{1}{e^x+2e^{-x}-3}=\dfrac{t}{\left(t-1\right)\left(t-2\right)}\) \(I=\int\limits^5_3\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}\) \(\int\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}=\ln\dfrac{t-2}{t-1}+C\) Hướng dẫn giải:Đặt \(t=e^x\) thì:
\(\dfrac{1}{e^x+2e^{-x}-3}=\dfrac{1}{t+2.\dfrac{1}{t}-3}=\dfrac{t}{t^2-3t+2}=\dfrac{t}{\left(t-1\right)\left(t-2\right)}\)
Ta lại có: \(\dfrac{1}{\left(t-1\right)\left(t-2\right)}=\dfrac{1}{t-2}-\dfrac{1}{t-1}\) nên
\(I=\int\limits^{\ln5}_{\ln3}\dfrac{\text{dx}}{e^x+2e^{-x}-3}=\int\limits^5_3\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}\)
Vì \(\int\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}=\int\dfrac{d\left(t-2\right)}{t-2}-\int\dfrac{d\left(t-1\right)}{t-1}=\ln\left|t-2\right|-\ln\left|t-1\right|+C=\ln\left|\dfrac{t-2}{t-1}\right|+C\)
nên \(I=\ln\left|\dfrac{t-2}{t-1}\right||^5_3=\ln\left|\dfrac{5-2}{5-1}\right|-\ln\left|\dfrac{3-2}{3-1}\right|=\ln\dfrac{3}{4}-\ln\dfrac{1}{2}\)
\(=\ln3-\ln4-\ln1+\ln2=\ln3-2\ln2+\ln2=\ln3-\ln2\)
Chú ý: \(\int\left(\dfrac{1}{t-2}-\dfrac{1}{t-1}\right)\text{dt}=\ln\left|\dfrac{t-2}{t-1}\right|+C\)