Đáp án:
Giải thích các bước giải:
Câu 22:
Có:
\(\begin{array}{l}
F'(x) = \frac{{f(x)}}{x} = \frac{{ - 4x}}{{4{x^4}}} = \frac{{ - 1}}{{{x^3}}}\\
\to f(x) = \frac{{ - 1}}{{{x^2}}} \to f'(x) = \frac{{ - 2x}}{{{x^4}}} = \frac{{ - 2}}{{{x^3}}}\\
\to \int {f'\left( x \right).\ln xdx = \int {\frac{{ - 2}}{{{x^3}}}.\ln xdx} }
\end{array}\)
Đặt :
\(\begin{array}{l}
u = \ln x \to du = \frac{{dx}}{x}\\
dv = \frac{1}{{{x^3}}}dx \to v = \frac{{ - 1}}{{2{x^2}}}\\
\to \int {\frac{{ - 2}}{{{x^3}}}.\ln xdx} = - 2\left[ {\frac{{ - \ln x}}{{2{x^2}}} + \int {\frac{1}{{2{x^3}}}} } \right]\\
= \frac{{\ln x}}{{{x^2}}} - \frac{{{x^{ - 2}}}}{{ - 2}} + C = \frac{{\ln x}}{{{x^2}}} + \frac{1}{{2{x^2}}} + C
\end{array}\)
