Question

Tính tích phân \(I=\int_1^e\dfrac{xln^2x}{\left(lnx+1\right)^2}dx\)

Answer 1

\(I=\int\limits^e_1x^2.ln^2x.\dfrac{1}{x\left(lnx+1\right)^2}dx\)

Đặt \(\left\{{}\begin{matrix}u=x^2ln^2x\\dv=\dfrac{1}{x\left(lnx+1\right)^2}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2x.lnx\left(lnx+1\right)\\v=-\dfrac{1}{lnx+1}\end{matrix}\right.\)

\(\Rightarrow I=-\dfrac{x^2ln^2x}{lnx+1}|^e_1+\int\limits^e_12x.lnxdx=-\dfrac{e^2}{2}+I_1\)

Xét \(I_1\), đặt \(\left\{{}\begin{matrix}u=lnx\\dv=2xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=x^2\end{matrix}\right.\)

\(\Rightarrow I_1=x^2lnx|^e_1-\int\limits^e_1xdx=...\)