\(I=\int _{0}^{e-1}\frac{x\ln \left(x+1\right)}{x^{2} +2x+1} dx \)
\(\int _{0}^{e-1}\frac{x\ln \left(x+1\right)}{x^{2} +2x+1} dx =\int _{0}^{e-1}\frac{\ln \left(x+1\right)}{x+1} dx -\int _{0}^{e-1}\frac{\ln \left(x+1\right)}{\left(x+1\right)^{2} } dx \)
Ta có:
- \(
I=\int _{0}^{e-1}\frac{\ln \left(x+1\right)}{x+1} dx =\frac{1}{2} \ln ^{2} \left(x+1\right)\left|\begin{array}{l} {e-1} \\ {0} \end{array}\right. =\frac{1}{2} \)
- \( J=\int _{0}^{e-1}\frac{\ln \left(x+1\right)}{\left(x+1\right)^{2} } dx \)
\( Đặt \left\{\begin{array}{l} {u=\ln \left(x+1\right)} \\ {dv=\frac{1}{\left(x+1\right)^{2} } dx} \end{array}\right.\) \(\Rightarrow \left\{\begin{array}{l} {du=\frac{1}{x+1} dx} \\ {v=-\frac{1}{x+1} } \end{array}\right.\)
\(I=-\frac{1}{x+1} x\ln \left(x+1\right)\left|\begin{array}{l} {e-1} \\ {0} \end{array}\right. +\int _{0}^{1}\frac{1}{\left(x+1\right)^{2} } xdx \)
\(=-\frac{1}{e} -\frac{1}{x+1} \left|\begin{array}{l} {e-1} \\ {0} \end{array}\right. =1-\frac{2}{e}\)
Vậy: \(\int _{0}^{e-1}\frac{x\ln \left(x+1\right)}{x^{2} +2x+1} dx =\frac{3}{2} -\frac{2}{e}\)