# Evaluate the integral. int sqrt x(x^3+x/2)dx

Question
Applications of integrals
Evaluate the integral.
$$\displaystyle\int\sqrt{{x}}{\left({x}^{{3}}+\frac{{x}}{{2}}\right)}{\left.{d}{x}\right.}$$

2020-10-22
Given That
$$\displaystyle\int\sqrt{{x}}{\left({x}^{{3}}+\frac{{x}}{{2}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left({x}^{{{3}+\frac{{1}}{{2}}}}+\frac{{{x}^{{\frac{{1}}{{2}}}}\cdot{x}}}{{2}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left({x}^{{\frac{{7}}{{2}}}}\right)}+\frac{{{x}^{{\frac{{3}}{{2}}}}}}{{2}}{)}{\left.{d}{x}\right.}$$
$$\displaystyle=\frac{{{x}^{{\frac{{7}}{{2}}}}+{1}}}{{\frac{{7}}{{2}}+{1}}}+\frac{{{x}^{{\frac{{3}}{{2}}}}+{1}}}{{\frac{{3}}{{2}}+{1}}}+{c}$$
$$\displaystyle\frac{{2}}{{9}}{x}^{{\frac{{9}}{{2}}}}+\frac{{1}}{{2}}\cdot\frac{{2}}{{5}}{x}^{{\frac{{5}}{{2}}}}+{c}$$
$$\displaystyle=\frac{{2}}{{9}}{x}^{{\frac{{9}}{{2}}}}+\frac{{1}}{{5}}{x}^{{\frac{{5}}{{2}}}}+{c}$$
$$\displaystyle\therefore\int\sqrt{{x}}{\left({x}^{{3}}+\frac{{x}}{{2}}\right)}{\left.{d}{x}\right.}=\frac{{2}}{{9}}{x}^{{\frac{{9}}{{2}}}}+\frac{{1}}{{5}}{x}^{{\frac{{5}}{{2}}}}+{c}$$

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