Tìm hàm số f(x) biết \(f\left( x \right) = \int {\frac{{5 + 4x}}{{{x^2}}}.lnxdx} .\)

Tìm hàm số f(x) biết \(f\left( x \right) = \int {\frac{{5 + 4x}}{{{x^2}}}.lnxdx} .\)
A. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x + 1} \right) + C\)
B. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x - 1} \right) + C\)
C. \(f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\ln x - \frac{5}{x}\)
D. \(f\left( x \right) = 2\ln x - \frac{5}{x}\left( {\ln x + 1} \right) + C\)

Ta có \(f\left( x \right) = \int {\frac{{5 + 4x}}{{{x^2}}}} \ln xdx = \int {\frac{{5\ln x}}{{{x^2}}}dx + \int {\frac{{4.\ln x}}{x}} dx = 2{{\ln }^2}x + \int {\frac{{5\ln x}}{{{x^2}}}} dx + C}\)
Đặt \(\left\{ {\begin{array}{*{20}{c}} {u = \ln x}\\ {dv = \frac{{dx}}{{{x^2}}}} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}} {du = \frac{{dx}}{x}}\\ {v = - \frac{1}{x}} \end{array}} \right.\)
\(\Rightarrow \int {\frac{{5\ln x}}{{{x^2}}}dx} = - \frac{{5\ln x}}{x} + 5.\int {\frac{{dx}}{{{x^2}}}} = - \frac{{5\ln x}}{x} - \frac{5}{x} + C\)
\(\Rightarrow f\left( x \right) = 2{\ln ^2}x - \frac{5}{x}\left( {\ln x + 1} \right) + C.\)