Cho \(\int\limits^3_1\frac{\text{d}x}{e^x-1}=\ln\left(ae^2+e+b\right)-2\). Tổng $a+b$ bằng
\(2\). \(-2\). \(4\). \(-4\). Hướng dẫn giải:Đặt \(t=e^x-1\) ⇒ \(\text{d}t=e^x\text{d}x\Rightarrow\text{d}x=\frac{\text{d}t}{e^x}=\frac{\text{d}t}{t+1}\)
Đổi cận: \(x|^3_1\Rightarrow t|^{e^3-1}_{e-1}\)
\(\int\limits^3_1\frac{1}{e^x-1}\text{d}x=\int\limits^{e^3-1}_{e-1}\frac{\text{d}t}{t\left(t+1\right)}\)
\(=\int\limits^{e^3-1}_{e-1}\left[\frac{1}{t}-\frac{1}{t+1}\right]\text{d}t\)
\(=\left[\ln\left|t\right|-\ln\left|t+1\right|\right]|^{e^3-1}_{e-1}\)
\(=\ln\frac{t}{t+1}|^{e^3-1}_{e-1}\)
\(=\ln\frac{e^3-1}{e^3}-\ln\frac{e-1}{e}\)
\(=\ln\left(\frac{e^3-1}{e^3}:\frac{e-1}{e}\right)\)
\(=\ln\frac{e^2+e+1}{e^2}\)
\(=\ln\left(e^2+e+1\right)-2\)
Vậy ta có:
\(\ln\left(e^2+e+1\right)-2=\ln\left(ae^2+e+b\right)-2\)
\(\Leftrightarrow e^2\left(1-a\right)=b-1\)
\(\Leftrightarrow\begin{cases}a=1\\b=1\end{cases}\)
\(\Rightarrow a+b=2\).
