Tính tích phân sau: I=∫1/(x^3(√x−√x-1))dx.

Question

Tính tích phân sau:

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

Answer 1

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

\(=\int _{1}^{2}\left(x^{\frac{-5}{2} } +\frac{\sqrt{x-1} }{x^{3} } \right){\rm d}x= \int _{1}^{2}x^{\frac{-5}{2} } dx+ \int _{1}^{2}\frac{\sqrt{x-1} }{x^{3} } {\rm d}x  \)
Tính \(\int _{1}^{2}x^{\frac{-5}{2} } {\rm d}x=\left. \frac{-2}{3.x^{\frac{3}{2} } } \right|_{1}^{2}  =\frac{-2}{3} .\left(\frac{1}{2\sqrt{2} } -1\right)=-\frac{1}{3\sqrt{2} } +\frac{2}{3} .\)

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

\( \int _{1}^{2}\frac{\sqrt{x-1} }{x^{3} } {\rm d}x =\int _{0}^{1}\frac{2t^{2} }{\left(t^{2} +1\right)^{3} } {\rm d}t .\) Đặt \(t=\tan u\Rightarrow {\rm d}t=\frac{1}{{\rm cos}^{2} u} {\rm d}u\)
\(\int _{0}^{1}\frac{2t^{2} }{\left(t^{2} +1\right)^{3} } {\rm d}t =\int _{0}^{\frac{\pi }{4} }\frac{2\tan ^{2} u}{\left(\frac{1}{\cos ^{2} u} \right)^{6} } .\frac{1}{{\rm cos}^{2} u} {\rm d}u\)

\(=\int _{0}^{\frac{\pi }{4} }2\sin ^{2} u.c{\rm os}^{2} u{\rm d}u= \frac{1}{2} \int _{0}^{\frac{\pi }{4} }\sin ^{2} 2u {\rm d}u \)
\(=\frac{1}{2} \int _{0}^{\frac{\pi }{4} }\frac{1-\cos 4u}{2}  {\rm d}u=\int _{0}^{\frac{\pi }{4} }\left(\frac{1}{4} -\frac{1}{2} {\rm cos}4u\right) {\rm d}u\)

\(=\left. \left(\frac{1}{4} u-\frac{1}{2} \frac{\sin 4u}{4} \right)\right|_{0}^{\frac{\pi }{4} } =\frac{\pi }{16} \). 
Vậy \( I =\int _{1}^{2}\frac{1}{x^{3} \left(\sqrt{x} -\sqrt{x-1} \right)} {\rm d}x =-\frac{1}{3\sqrt{2} } +\frac{2}{3} +\frac{\pi }{16} \).