# Right, or wrong? Say which for each formula and give a brief reason for each answer. int(2x + 1)2 dx =(2x + 1)3/ 3 + C

Applications of integrals
Right, or wrong? Say which for each formula and give a brief reason for each answer. $$\displaystyle\int{\left({2}{x}+{1}\right)}{2}{\left.{d}{x}\right.}={\left({2}{x}+{1}\right)}\frac{{3}}{{3}}+{C}$$

2021-02-21
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: $$\displaystyle\int{\left({2}{x}+{1}\right)}^{{2}}{\left.{d}{x}\right.}=\frac{{{\left({2}{x}+{1}\right)}^{{3}}}}{{6}}+{C}$$