Cho \(\int {{{\left( {\frac{x}{{x + 1}}} \right)}^2}dx = mx + n\ln \left| {x + 1} \right| + \frac{p}{{x + 1}} + C} \).

* Hướng dẫn giải

\(\begin{array}{l}
\int {{{\left( {\frac{x}{{x + 1}}} \right)}^2}dx = \int {\frac{{{x^2}}}{{{{\left( {x + 1} \right)}^2}}}dx = \int {\left( {1 - \frac{{2x + 1}}{{{{\left( {x + 1} \right)}^2}}}} \right)} } } dx = \int {\left( {1 - \frac{{2x + 2}}{{{{\left( {x + 1} \right)}^2}}} + \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx\\
 = \int {dx - \int {\frac{{2x + 2}}{{{{\left( {x + 1} \right)}^2}}}} } dx + \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} dx = x - \int {\frac{{d\left( {{{\left( {x + 1} \right)}^2}} \right)}}{{{{\left( {x + 1} \right)}^2}}}}  - \frac{1}{{x + 1}} + C\\
 = x - \ln \left| {{{\left( {x + 1} \right)}^2}} \right| - \frac{1}{{x + 1}} + C = x - 2\ln \left| {x + 1} \right| - \frac{1}{{x + 1}} + C\\
 \Rightarrow m = 1,n =  - 2,p =  - 1 \Rightarrow m + n + p =  - 2
\end{array}\)