Tính tích phân sau: I=∫(2(√x^2+3)−x^2)/x^3(√x^2+3)dx

Question

Tính tích phân sau: 

\( I =\int _{1}^{2}\frac{2\sqrt{x^{2} +3} -x^{2} }{x^{3} \sqrt{x^{2} +3} } {\rm d}x  \)

Answer 1

\(I =\int _{1}^{2}\frac{2\sqrt{x^{2} +3} -x^{2} }{x^{3} \sqrt{x^{2} +3} } {\rm d}x\) \(=\int _{1}^{2}\left(\frac{2}{x^{3} } -\frac{x^{2} }{x^{3} \sqrt{x^{2} +3} } \right){\rm d}x\)

\(= \int _{1}^{2}\frac{2}{x^{3} } {\rm d}x- \int _{1}^{2}\frac{x^{2} }{x^{3} \sqrt{x^{2} +3} } {\rm d}x  \)
+ Tính \(\int _{1}^{2}\frac{2}{x^{3} } {\rm d}x =\left. -\frac{1}{x^{2} } \right|_{1}^{2} =\frac{3}{4} \).

 + Tính \(\int _{1}^{2}\frac{x^{2} }{x^{3} \sqrt{x^{2} +3} } {\rm d}x =\int _{1}^{2}\frac{x}{x^{2} \sqrt{x^{2} +3} } {\rm d}x . \)

Đặt \(\sqrt{x^{2} +3} =t\Rightarrow x^{2} +3=t^{2} \Rightarrow x{\rm d}x=t{\rm d}t.\)

 Ta có: \(\int _{1}^{2}\frac{x}{x^{2} \sqrt{x^{2} +3} } {\rm d}x \)\( =\int _{2}^{\sqrt{7} }\frac{t}{\left(t^{2} -3\right)t} {\rm d}t \)

\(=\int _{2}^{\sqrt{7} }\frac{1}{t^{2} -3} {\rm d}t =\frac{1}{2\sqrt{3} } .\int _{2}^{\sqrt{7} }\left(\frac{1}{t-\sqrt{3} } -\frac{1}{t+\sqrt{3} } \right){\rm d}t \)
\(=\left. \frac{1}{6} .\left(\ln \left|t-\sqrt{3} \right|-\ln \left|t+\sqrt{3} \right|\right)\right|_{2}^{\sqrt{7} }\)

\( = \frac{1}{2\sqrt{3} } .\left(\ln \frac{5-\sqrt{21} }{2} +\ln \left(7+4\sqrt{3} \right)\right). \)
Vậy\( I =\int _{1}^{2}\frac{2\sqrt{x^{2} +3} -x^{2} }{x^{3} \sqrt{x^{2} +3} } {\rm d}x =\frac{3}{4} -\frac{1}{2\sqrt{3} } .\left(\ln \frac{5-\sqrt{21} }{2} +\ln \left(7+4\sqrt{3} \right)\right).\)