Lời giải của Tự Học 365

$\begin{array}{l}f'(x) = \frac{1}{{{x^2} - 1}} \Rightarrow \int {f'(x)dx}  = \int {\frac{1}{{{x^2} - 1}}dx}  \Rightarrow f(x) = \int {\frac{1}{{(x - 1)(x + 1)}}dx}  = \frac{1}{2}\int {\frac{1}{{x - 1}}dx + } \frac{1}{2}\int {\frac{1}{{x + 1}}dx = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C} \\ \Rightarrow f(x) = \left\{ \begin{array}{l}\frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_1},\left[ \begin{array}{l}x <  - 1\\x > 1\end{array} \right.\\\frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + {C_2}, - 1 < x < 1\end{array} \right.\end{array}$

Ta có:

$f(-3)+f(3)=0\Leftrightarrow \frac{1}{2}\ln 2+{{C}_{1}}+\frac{1}{2}\ln \frac{1}{2}+{{C}_{1}}=0\Leftrightarrow 2{{C}_{1}}=0\Leftrightarrow {{C}_{1}}=0$

$f\left( -\frac{1}{2} \right)+f\left( \frac{1}{2} \right)=2\Leftrightarrow \frac{1}{2}\ln 3+{{C}_{2}}+\frac{1}{2}\ln \frac{1}{3}+{{C}_{2}}=2\Leftrightarrow {{C}_{2}}=1$

$P=f(0)+f(4)=\left( \frac{1}{2}\ln \left| \frac{0-1}{0+1} \right|+1 \right)+\left( \frac{1}{2}\ln \left| \frac{4-1}{4+1} \right| \right)=\frac{1}{2}\ln \frac{3}{5}+1$

Đáp án cần chọn là: a