Biết $\Large F(x)$ là một nguyên hàm của hàm số $\Large f(x)=\dfrac{{{

Biết $\Large F(x)$ là một nguyên hàm của hàm số $\Large f(x)=\dfrac{{{e}^{x}}+{{e}^{-x}}+3}{{{e}^{x}}+{{e}^{-x}}+2}$ và $\Large F(0)+F(\ln 2)=-\dfrac{13}{30}$. Tính $\Large F(\ln 4)$.

• A.

  $\Large F(\ln 4)=\dfrac{1}{2}\ln 2$

• B.

  $\Large F(\ln 4)=\dfrac{5}{2}\ln 2$

• C.

  $\Large F(\ln 4)=-\dfrac{1}{2}\ln 2$

• D.

  $\Large F(\ln 4)=\dfrac{3}{2}\ln 2$

Ta có: $\Large F(x)=\int{\dfrac{{{e}^{x}}+{{e}^{-x}}+3}{{{e}^{x}}+{{e}^{-x}}+2}dx=\int{\dfrac{{{e}^{x}}+\dfrac{1}{{{e}^{x}}}+3}{{{e}^{x}}+\dfrac{1}{{{e}^{x}}}+2}dx=\int{\dfrac{{{e}^{2x}}+3{{e}^{x}}+1}{{{e}^{2x}}+2{{e}^{x}}+1}dx}}}$ $\Large =\int{\left[ 1+\dfrac{{{e}^{x}}}{{{({{e}^{x}}+1)}^{2}}} \right]dx=\int{dx+\int{\dfrac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}dx=x+\int{\dfrac{{{e}^{x}}dx}{{{({{e}^{x}}+1)}^{2}}}}}}}$

Đặt $\Large t={{e}^{x}}+1\Rightarrow dt={{e}^{x}}dx$, khi đó $\Large \int{\dfrac{{{e}^{x}}dx}{{{({{e}^{x}}+1)}^{2}}}=\int{\dfrac{dt}{{{t}^{2}}}=-\dfrac{1}{t}+C=-\dfrac{1}{{{e}^{x}}+1}+C}}$

Vậy $\Large F(x)=x-\dfrac{1}{{{e}^{x}}+1}+C$, khi đó : $\Large F(0)+F(\ln 2)=-\dfrac{13}{30}\Leftrightarrow -\dfrac{1}{2}+C+\ln 2-\dfrac{1}{3}+C=-\dfrac{13}{30}$ $\Large \Leftrightarrow C=\dfrac{1}{5}-\dfrac{1}{2}\ln 2$

Suy ra $\Large F(\ln 4)=\ln 4-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{2}\ln 2=\dfrac{3}{2}\ln 2\to $ đáp án D