Lời giải:
Đặt \(x=\sqrt{3}\tan t(t\in (0; \frac{\pi}{2}))\)
\(\Rightarrow \sqrt{9+3x^2}=\sqrt{9+9\tan ^2t}=\sqrt{\frac{9}{\cos ^2t}}=\frac{3}{\cos t}\)
Khi đó \(I=\int \frac{3d(\sqrt{3}\tan t)}{3\cos t.\tan ^2t}=\int \frac{d(\sqrt{3}\tan t)}{\cos t.\tan ^2t}\)
\(=\int \frac{\sqrt{3}dt}{\cos ^3t\tan ^2t}=\sqrt{3}\int \frac{dt}{\cos ^3.\frac{\sin ^2t}{\cos ^2t}}\)
\(=\sqrt{3}\int \frac{dt}{\cos t\sin ^2t}\)
Đặt \(\left\{\begin{matrix} u=\frac{1}{\cos t}\\ dv=\frac{dt}{\sin ^2t}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{\sin t}{\cos ^2t}dt\\ v=-\cot t\end{matrix}\right.\)
Suy ra \(I=\sqrt{3}(\frac{-\cot t}{\cos t}+\int \frac{\cot t\sin t}{\cos ^2t}dt)\)
\(=\sqrt{3}(\frac{-\cot t}{\cos t}+\int \frac{dt}{\cos t})\)
\(=\sqrt{3}(\frac{-\cot t}{\cos t}+\int \frac{d(\sin t)}{1-\sin ^2t})\)
Phân tích:
\(\int \frac{d(\sin t)}{1-\sin ^2t}=\int \frac{dk}{1-k^2}=\frac{1}{2}\int \frac{dk}{1-k}+\frac{1}{2}\int \frac{dk}{1+k}=\frac{1}{2}\ln |k+1|-\frac{1}{2}\ln |1-k|+c\)
\(=\frac{1}{2}\ln |\frac{\sin t+1}{\sin t-1}|+c\)
Vậy \(I=\sqrt{3}(\frac{\cot t}{\cos t}+\frac{1}{2}\ln |\frac{\sin t+1}{\sin t-1}|)+c\)
