​ Cho tích phân I=∫21(x2−1)lnxdx . Khi đó: ​

Question

Cho tích phân \(I=\int _{1}^{2}\left(x^{2} -1\right)\ln x{\rm d}x \). Khi đó:

 A. \(I=\frac{2\ln 2+6}{9} \). B. \(I=\frac{6\ln 2+2}{9}\) . C.\( I=\frac{2\ln 2-6}{9}\) . D. \(I=\frac{6\ln 2-2}{9} . \)

Answer 1

Chọn B

 Đặt :  \(\left\{\begin{array}{l} {u=\ln x} \\ {{\rm dv}=\left(x^{2} -1\right){\rm d}x} \end{array}\right. \Rightarrow \left\{\begin{array}{l} {{\rm d}u=\frac{1}{x} {\rm d}x} \\ {v=\frac{x^{3} }{3} -x} \end{array}\right.  .\)

 Khi đó: 
\(I=\left. \left(\frac{x^{3} }{3} -x\right)\ln x\right|_{1}^{2} -\int _{1}^{2}\frac{1}{x} \left(\frac{x^{3} }{3} -x\right){\rm d}x \)

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

\(=\frac{2}{3} \ln 2-\left. \left(\frac{x^{3} }{9} -x\right)\right|_{1}^{2} \, \, =\frac{2}{3} \ln 2+\frac{2}{9} =\frac{6\ln 2+2}{9} .\)