Question

$\int \sqrt{x}(3x^2+2x+3)dx$ is equal to (where $C$ is the constant of integration)

• A. $\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$
• B. $\frac{6}{7}{x}^{\frac{7}{3}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$
• C. $\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{3}}+2x^{\frac{3}{2}}+C$
• D. $\frac{6}{7}{x}^{\frac{7}{4}}-\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$

Answer 1

The correct option is A $\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$

$I=\int \sqrt{x}(3x^2+2x+3)dx=\int (3x^{\frac{5}{2}}+2x^{\frac{3}{2}}+3x^{\frac{1}{2}})dx$
By taking the terms separately, we get
$=3\int x^{\frac{5}{2}}dx+2\int x^{\frac{3}{2}}dx+3\int x^{\frac{1}{2}}dx=3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$
$\Rightarrow I=\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$