Tính tích phân sau: I=∫((x2−2x+3−√x^2+1))/xdx.

Question

Tính tích phân sau: 

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

Answer 1

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

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

\(\Rightarrow {\rm d}x=\frac{tdt}{x} \)

 Đổi cận: \(x=1\Rightarrow t=\sqrt{2} ;\, \, x=2\Rightarrow t=\sqrt{5} \)

 Do đó \(\int _{1}^{2}\frac{\sqrt{x^{} +1} }{x} dx\) \(=\int _{\sqrt{2} }^{\sqrt{5} }\frac{t^{2} }{t^{2} -1}  dt\)

\(=\int _{\sqrt{2} }^{\sqrt{5} }\left(1+\frac{1}{t^{2} -1} \right) dt=\int _{\sqrt{2} }^{\sqrt{5} }\left[1+\frac{1}{2} \left(\frac{1}{t-1} -\frac{1}{t+1} \right)\right] dt\)
\(=\left(t+\frac{1}{2} \ln \left|\frac{t-1}{t+1} \right|\right)\left|\begin{array}{l} {\sqrt{5} } \\ {\sqrt{2} } \end{array}\right. \)\(=\sqrt{5} -\sqrt{2} +\frac{1}{2} \ln \frac{\left(\sqrt{5} -1\right)\left(\sqrt{2} +1\right)}{\left(\sqrt{5} +1\right)\left(\sqrt{2} -1\right)}  \)
Vậy \(I =\frac{-1}{2} +3\ln 2-\sqrt{5} +\sqrt{2} -\frac{1}{2} \ln \frac{\left(\sqrt{5} -1\right)\left(\sqrt{2} +1\right)}{\left(\sqrt{5} +1\right)\left(\sqrt{2} -1\right)} .\)