Đáp án:

$i) I=\dfrac{32}{3}$

$j) K=\dfrac{1}{4}\ln\dfrac{5}{3}$

$k) L=\dfrac{\pi}{24}$

$l) M=\pi$

Giải thích các bước giải:

$i) I=\int\limits^{\ln8}_0\dfrac{e^x+e^{2x}}{\sqrt{e^x+8}}\mathrm{d}x$

Đặt $t=\sqrt{e^x+8}\to t^2=e^x+8 \to \mathrm{d}t=\dfrac{e^x}{2\sqrt{e^x+8}}\mathrm{d}x \to e^x\mathrm{d}x=2t\mathrm{d}t$

Đổi cận: $\begin{cases}x=0 \to t=3\\x=\ln8 \to t=4\end{cases}$

$\to I=\int\limits^{\ln8}_0\dfrac{e^x(e^x+1)\mathrm{d}x}{\sqrt{e^x+8}}\\=\int\limits^4_3\dfrac{(t^2-7).2t\mathrm{d}t}{t}\\=2\int\limits^4_3(t^2-7)\mathrm{d}t\\=2\Bigg(\dfrac{t^3}{3}-7t)\Bigg)\Bigg|^4_3\\=2\Bigg(\dfrac{4^3}{3}-7.4-\dfrac{3^3}{3}+3.7\Bigg)\\=\dfrac{32}{3}$

$j) K=\int\limits^{2\sqrt{3}}_\sqrt{5}\dfrac{\mathrm{d}x}{x\sqrt{x^2+4}}$

Đặt $t=\sqrt{x^2+4}\to x^2=t^2-4 \to \mathrm{d}t=\dfrac{x}{\sqrt{x^2+4}}\mathrm{d}x \to x\mathrm{d}x=t\mathrm{d}t$

Đổi cận: $\begin{cases}x=\sqrt{5} \to t=3\\x=2\sqrt{3} \to t=4\end{cases}$

$\to K=\int\limits^4_3\dfrac{t\mathrm{d}t}{x^2.t}\\=\int\limits^4_3\dfrac{1}{t^2-4}\mathrm{d}t\\=\dfrac{1}{4}\int\limits^4_3\Bigg(\dfrac{1}{t-2}-\dfrac{1}{t+2}\Bigg)\mathrm{d}t\\=\dfrac{1}{4}\ln\Big|\dfrac{t-2}{t+2}\Big|\Bigg|^4_3\\=\dfrac{1}{4}\Bigg(\ln\dfrac{2}{6}-\ln\dfrac{1}{5}\Bigg)\\=\dfrac{1}{4}\ln\dfrac{5}{3}$

$k) L=\int\limits^{2\sqrt{3}}_2\dfrac{1}{x^2+4}\mathrm{d}x$

Đặt $x=2\tan t \hspace{1,5cm}t\in \Bigg(-\dfrac{\pi}{2};\,\dfrac{\pi}{2}\Bigg)\\ \to \mathrm{d}x=2(\tan^2t+1)\mathrm{d}t$

Đổi cận: $\begin{cases}x=\sqrt{2} \to t=\dfrac{\pi}{4}\\x=2\sqrt{3} \to t=\dfrac{\pi}{3}\end{cases}$

$\to L=\int\limits^\dfrac{\pi}{3}_\dfrac{\pi}{4}\dfrac{2(\tan^2t+1)\mathrm{d}t}{4\tan^2t+4}\\=\dfrac{1}{2}\int\limits^\dfrac{\pi}{3}_\dfrac{\pi}{4}\mathrm{d}t\\=\dfrac{1}{2}t\Bigg|^\dfrac{\pi}{3}_\dfrac{\pi}{4}\\=\dfrac{\pi}{6}-\dfrac{\pi}{8}\\=\dfrac{\pi}{24}$

$l) M=\int\limits^2_0 \sqrt{4-x^2}\mathrm{d}x$

Đặt $x=2\sin t  \to \mathrm{d}x=2\cos t \mathrm{d}t$

Đổi cận: $\begin{cases}x=0 \to t=0\\x=2 \to t=\dfrac{\pi}{2}\end{cases}$

$\to M=\int\limits^\dfrac{\pi}{2}_0 \sqrt{4-4\sin^2t}.2\cos t \mathrm{d}t\\=4\int\limits^\dfrac{\pi}{2}_0 \cos^2t \mathrm{d}t\\=2\int\limits^\dfrac{\pi}{2}_0(\cos 2t +1)\mathrm{d}t\\=(\sin 2t+2t)\Bigg|^\dfrac{\pi}{2}_0\\=\pi$