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