Đáp án:
\( \dfrac{{{x^2}}}{2} + x + 2\ln \left| {\dfrac{{x - 1}}{{x + 3}}} \right| + C\)
Giải thích các bước giải:
\(\begin{array}{l}\int {\dfrac{{{{\left( {x + 1} \right)}^3}}}{{{x^2} + 2x - 3}}dx} = \int {\dfrac{{{x^3} + 3{x^2} + 3x + 1}}{{{x^2} + 2x - 3}}dx} \\ = \int {\left( {x + 1 + \dfrac{{4x + 4}}{{{x^2} + 2x - 3}}} \right)dx} = \int {\left( {x + 1 + \dfrac{{4x + 4}}{{\left( {x - 1} \right)\left( {x + 3} \right)}}} \right)dx} \end{array}\)
Xét \(\dfrac{{4x + 4}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = \dfrac{a}{{x - 1}} + \dfrac{b}{{x + 3}}\) ta có:
\(\dfrac{a}{{x - 1}} + \dfrac{b}{{x + 3}} = \dfrac{{a\left( {x + 3} \right) + b\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = \dfrac{{\left( {a + b} \right)x + 3a - b}}{{\left( {x - 1} \right)\left( {x + 3} \right)}}\)
\( \Rightarrow \left\{ \begin{array}{l}a + b = 4\\3a - b = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 2\end{array} \right.\)
\( \Rightarrow \dfrac{{4x + 4}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = \dfrac{2}{{x - 1}} + \dfrac{2}{{x + 3}}\)
\(\begin{array}{l} \Rightarrow I = \int {\left( {x + 1 + \dfrac{2}{{x - 1}} + \dfrac{2}{{x + 3}}} \right)dx} \\ = \dfrac{{{x^2}}}{2} + x + 2\ln \left| {x - 1} \right| + 2\ln \left| {x + 3} \right| + C\\ = \dfrac{{{x^2}}}{2} + x + 2\ln \left| {\dfrac{{x - 1}}{{x + 3}}} \right| + C\end{array}\)
