Step 1

Consider the integral, \(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}\)

Step 2

To solve the given integrals,

\(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\int{\left({4}{x}^{{2}}+{1}+{4}{x}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={4}\int{x}^{{2}}{\left.{d}{x}\right.}+\int{1}\cdot{\left.{d}{x}\right.}+{4}\int{x}{\left.{d}{x}\right.}\)

\(\displaystyle={4}{\left(\frac{{{x}^{{3}}}}{{3}}\right)}+{x}+{4}{\left(\frac{{{x}^{{2}}}}{{2}}\right)}+{C}\)

\(\displaystyle=\frac{{{4}{x}^{{3}}}}{{3}}+{2}{x}^{{2}}+{x}+{C}\)

Step 3

\(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{{3}\cdot{2}}}+{C}\)

\(\displaystyle=\frac{{{8}{x}^{{3}}+{1}+{6}{x}{\left({2}{x}+{1}\right)}}}{{6}}+{C}\)

\(\displaystyle=\frac{{{8}{x}^{{3}}+{1}+{12}{x}^{{2}}+{6}{x}}}{{6}}+{C}\)

\(\displaystyle=\frac{{{8}{x}^{{3}}}}{{6}}+\frac{{{12}{x}^{{2}}}}{{6}}+{6}\frac{{x}}{{6}}+\frac{{1}}{{6}}+{C}\)

\(\displaystyle=\frac{{{4}{x}^{{3}}}}{{3}}+{2}{x}^{{2}}+{x}+\frac{{1}}{{6}}+{C}\)

Hence the given integral: \(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{3}}+{C}\) is wrong.

The right answer is: \(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{6}}+{C}\)

Consider the integral, \(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}\)

Step 2

To solve the given integrals,

\(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\int{\left({4}{x}^{{2}}+{1}+{4}{x}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={4}\int{x}^{{2}}{\left.{d}{x}\right.}+\int{1}\cdot{\left.{d}{x}\right.}+{4}\int{x}{\left.{d}{x}\right.}\)

\(\displaystyle={4}{\left(\frac{{{x}^{{3}}}}{{3}}\right)}+{x}+{4}{\left(\frac{{{x}^{{2}}}}{{2}}\right)}+{C}\)

\(\displaystyle=\frac{{{4}{x}^{{3}}}}{{3}}+{2}{x}^{{2}}+{x}+{C}\)

Step 3

\(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{{3}\cdot{2}}}+{C}\)

\(\displaystyle=\frac{{{8}{x}^{{3}}+{1}+{6}{x}{\left({2}{x}+{1}\right)}}}{{6}}+{C}\)

\(\displaystyle=\frac{{{8}{x}^{{3}}+{1}+{12}{x}^{{2}}+{6}{x}}}{{6}}+{C}\)

\(\displaystyle=\frac{{{8}{x}^{{3}}}}{{6}}+\frac{{{12}{x}^{{2}}}}{{6}}+{6}\frac{{x}}{{6}}+\frac{{1}}{{6}}+{C}\)

\(\displaystyle=\frac{{{4}{x}^{{3}}}}{{3}}+{2}{x}^{{2}}+{x}+\frac{{1}}{{6}}+{C}\)

Hence the given integral: \(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{3}}+{C}\) is wrong.

The right answer is: \(\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{6}}+{C}\)