Tìm nguyên hàm của hàm số sau: ∫3+lnx/(x+1)2dx

Question

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

Answer 1

Ta có: \(\int \frac{3+\ln x}{\left(x+1\right)^{2} } {\rm d}x =\int \frac{3}{\left(x+1\right)^{2} } {\rm d}x +\int \frac{\ln x}{\left(x+1\right)^{2} } {\rm d}x  =\frac{-3}{x+1} +J.\)

 Tính \(J=\int \frac{\ln x}{\left(x+1\right)^{2} } {\rm d}x .\)

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

 Khi đó: \(J=-\frac{\ln x}{x+1} +\int \frac{{\rm d}x}{x\left(x+1\right)}  =-\frac{\ln x}{x+1} +\ln \left(\frac{x}{x+1} \right)+C.\)

 Vậy: \( \int \frac{3+\ln x}{\left(x+1\right)^{2} } {\rm d}x =\frac{-3}{x+1} +J=\frac{-3}{x+1} -\frac{\ln x}{x+1} +\ln \left(\frac{x}{x+1} \right)+C.\)