# Find the integral, full solution \int (2x+3)^{2}dx

Find the integral, full solution
$$\displaystyle\int{\left({2}{x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}$$

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Mary Herrera
Step 1
Given integral,
$$\displaystyle\int{\left({2}{x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}$$
Step 2
$$\displaystyle\int{\left({2}{x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}=\int{\left({4}{x}^{{{2}}}+{12}{x}+{9}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{4}{x}^{{{2}}}{\left.{d}{x}\right.}+\int{12}{x}{\left.{d}{x}\right.}+\int{9}{\left.{d}{x}\right.}$$
$$\displaystyle={4}\int{x}^{{{2}}}{\left.{d}{x}\right.}+{12}\int{x}{\left.{d}{x}\right.}+{9}\int{\left.{d}{x}\right.}$$
$$\displaystyle={4}{\frac{{{x}^{{{3}}}}}{{{3}}}}+{12}{\frac{{{x}^{{{2}}}}}{{{2}}}}+{9}{x}+{C}$$ Since $$\displaystyle\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C},\ \ \ {n}\ne{1}$$
$$\displaystyle={\frac{{{4}{x}^{{{3}}}}}{{{3}}}}+{6}{x}^{{{2}}}+{9}{x}+{C}$$
###### Not exactly what youâ€™re looking for?
Raymond Foley
Calculate:
$$\displaystyle\int{\left({2}{x}+{3}\right)}^{{{2}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}\int{u}^{{{2}}}{d}{u}$$
$$\displaystyle\int{u}^{{{2}}}{d}{u}$$
$$\displaystyle={\frac{{{u}^{{{3}}}}}{{{3}}}}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}\int{u}^{{{2}}}{d}{u}$$
$$\displaystyle={\frac{{{u}^{{{3}}}}}{{{6}}}}$$
$$\displaystyle={\frac{{{\left({2}{x}+{3}\right)}^{{{3}}}}}{{{6}}}}$$
$$\displaystyle={\frac{{{\left({2}{x}+{3}\right)}^{{{3}}}}}{{{6}}}}+{C}$$
Let's simplify:
$$\displaystyle{\frac{{{4}{x}^{{{3}}}}}{{{3}}}}+{6}{x}^{{{2}}}+{9}{x}+{C}$$
karton

$$\int(2x+3)^{2}dx \\\int 4x^{2}+12x+9dx \\\int 4x^{2}dx+\int 12xdx+\int 9dx \\\frac{4x^{3}}{3}+6x^{2}+9x \\\text{Add C} \\\text{Answer:} \\\frac{4x^{3}}{3}+6x^{2}+9x+C$$