Tích phân \(\int\limits^{\ln5}_0\dfrac{e^x\sqrt{e^x-1}}{e^x+3}\text{d}x\) bằng
\(4-\pi\).\(4+\pi\).\(2-\pi\).\(2+\pi\).Hướng dẫn giải:Đặt \(t=\sqrt{e^x-1}\Rightarrow t^2=e^x-1\Rightarrow2t\text{d}t=e^x\text{d}x\)
Đổi cận: \(x=0\Rightarrow t=0;x=\ln5\Rightarrow t=2\).
\(\int\limits^{\ln5}_0\dfrac{e^x\sqrt{e^x-1}}{e^x+3}\text{d}x=\)\(\int\limits^2_0\dfrac{2t^2}{t^2+4}\text{d}t=\int\limits^2_0\dfrac{2\left(t^2+4\right)-8}{t^2+4}\text{d}t=2\left(\int\limits^2_0\text{d}t-4\int\limits^2_0\dfrac{1}{t^2+4}\text{d}t\right)=4-8\int\limits^2_0\dfrac{\text{d}t}{t^2+4}\)
Để tính \(\int\limits^2_0\dfrac{1}{t^2+4}\text{d}t\) ta đặt \(t=2\tan u\Rightarrow\text{d}=2\left(1+\tan^2u\right)\text{d}u\). Đổi cận: \(t=0\Rightarrow u=0;t=2\Rightarrow u=\dfrac{\pi}{4}\) nên
\(\int\limits^2_0\dfrac{1}{t^2+4}\text{d}t=2\int\limits^{\frac{\pi}{4}}_0\dfrac{1}{4}du=\dfrac{\pi}{8}\Rightarrow\int\limits^{\ln5}_0\dfrac{e^x\sqrt{e^x-1}}{e^x+3}=4-8.\dfrac{\pi}{8}=4-\pi.\)