Answer
\[\begin{align}
& \text{a}\text{. }\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+C \\
& \text{b}\text{. }\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+C \\
& \text{c}\text{.The antiderivatives differ by a constant}\text{.} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \text{a}\text{. }\int{\frac{{{x}^{2}}}{x+1}}dx \\
& \text{Let }u=x+1,\text{ }x=u-1,\text{ }dx=du \\
& \int{\frac{{{x}^{2}}}{x+1}}dx=\int{\frac{{{\left( u-1 \right)}^{2}}}{u}}du \\
& \text{ }=\int{\frac{{{u}^{2}}-2u+1}{u}}du \\
& \text{ }=\int{\left( u-2+\frac{1}{u} \right)}du \\
& \text{Integrating} \\
& =\frac{1}{2}{{u}^{2}}-2u+\ln \left| u \right|+C \\
& \text{Back-substitute }u=x+1 \\
& =\frac{1}{2}{{\left( x+1 \right)}^{2}}-2\left( x+1 \right)+\ln \left| x+1 \right|+C \\
& \text{Simplify and combine constants} \\
& =\frac{1}{2}\left( {{x}^{2}}+2x+1 \right)-2x-2+\ln \left| x+1 \right|+C \\
& =\frac{1}{2}{{x}^{2}}+x+\frac{1}{2}-2x-2+\ln \left| x+1 \right|+C \\
& =\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+{{C}_{1}} \\
& \\
& \text{b}\text{. }\int{\frac{{{x}^{2}}}{x+1}}dx \\
& \text{By the long division} \\
& \int{\frac{{{x}^{2}}}{x+1}}dx=\int{\left( x-1+\frac{1}{x+1} \right)}dx \\
& \text{Integrating} \\
& =\frac{1}{2}{{x}^{2}}-x+\ln \left| x+1 \right|+C \\
& \\
& \text{c}\text{.The antiderivatives differ by a constant}\text{.} \\
\end{align}\]