Cho hàm số $\Large f(x)$ xác định trên $\Large R\backslash \left[ -1;1

Cho hàm số $\Large f(x)$ xác định trên $\Large R\backslash \left[ -1;1 \right]$ và thỏa mãn $\Large {f}'(x)=\dfrac{1}{{{x}^{2}}-1}$ . Biết rằng $\Large f(-3)+f(3)=0$ và $\Large f\left( -\dfrac{1}{2} \right)+f\left( \dfrac{1}{2} \right)=2$ . Tính $\Large T=f(-2)+f(0)+f(4)$

• A.

  $\Large T=1+\ln \dfrac{9}{5}$

• B.

  $\Large T=1+\ln \dfrac{6}{5}$

• C.

  $\Large T=1+\dfrac{1}{2}\ln \dfrac{9}{5}$

• D.

  $\Large T=1+\dfrac{1}{2}\ln \dfrac{6}{5}$

Ta có $\Large f(x)=\int{\frac{1}{{{x}^{2}}-1}dx=\frac{1}{2}\int{\left( \frac{1}{x-1}-\frac{1}{x+1} \right)dx=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+C}}$

Với $\Large x\in \left( -\infty ;-1 \right)$ ta có $\Large f(x)=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+{{C}_{1}}$

Với $\Large x\in \left( 1;+\infty  \right)$ ta có $\Large f(x)=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+{{C}_{3}}$

Mà $\Large f(-3)+f(3)=0\Leftrightarrow \frac{1}{2}\ln \left| \frac{-3-1}{-3+1} \right|+{{C}_{1}}+\frac{1}{2}\ln \left| \frac{3-1}{3+1} \right|+{{C}_{3}}=0$

$\Large \Leftrightarrow \frac{1}{2}\ln 2+{{C}_{1}}+\frac{1}{2}\ln \frac{1}{2}+{{C}_{3}}=0\Leftrightarrow {{C}_{1}}+{{C}_{3}}=0$

Do đó $\Large f(-2)=\frac{1}{2}\ln 3+{{C}_{1}};f(4)=\frac{1}{2}\ln \frac{3}{5}+{{C}_{3}}$

Với $\Large x\in \left( -1;1 \right)$ ta có $\Large f(x)=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+{{C}_{2}}$

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

$\Large \Leftrightarrow \frac{1}{2}\ln 3+{{C}_{2}}+\frac{1}{2}\ln \frac{1}{3}+{{C}_{2}}=2\Leftrightarrow {{C}_{2}}=1$

Do đó với $\Large x\in \left( -1;1 \right),f(x)=\frac{1}{2}\ln \left| \frac{x-1}{x+1} \right|+1\Rightarrow f(0)=1$

Vậy $\Large T=f(-2)+f(0)+f(4)=1+\frac{1}{2}\ln \frac{9}{5}$

Chọn đáp án C