Đáp án:
\(\dfrac{{3x + 1}}{{\left( {1 + {x^2}} \right)\left( {x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{1}{{x + 1}} + \dfrac{{x - {x^3}}}{{1 + {x^2}}}.\left( {\dfrac{1}{{1 + 2x + {x^2}}} - \dfrac{1}{{{x^2} - 1}}} \right)\\
= \dfrac{1}{{x + 1}} + \dfrac{{x\left( {1 - {x^2}} \right)}}{{1 + {x^2}}}.\left( {\dfrac{1}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right)\\
= \dfrac{1}{{x + 1}} + \dfrac{{x\left( {1 - {x^2}} \right)}}{{1 + {x^2}}}.\left[ {\dfrac{{x - 1 - x - 1}}{{\left( {{x^2} - 1} \right)\left( {x + 1} \right)}}} \right]\\
= \dfrac{1}{{x + 1}} + \dfrac{{x\left( {1 - {x^2}} \right)}}{{1 + {x^2}}}.\left[ {\dfrac{{ - 2}}{{\left( {{x^2} - 1} \right)\left( {x + 1} \right)}}} \right]\\
= \dfrac{1}{{x + 1}} + \dfrac{{x\left( {1 - {x^2}} \right)}}{{1 + {x^2}}}.\left[ {\dfrac{2}{{\left( {1 - {x^2}} \right)\left( {x + 1} \right)}}} \right]\\
= \dfrac{1}{{x + 1}} + \dfrac{{2x}}{{\left( {1 + {x^2}} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x + 1 + 2x}}{{\left( {1 + {x^2}} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x + 1}}{{\left( {1 + {x^2}} \right)\left( {x + 1} \right)}}
\end{array}\)
