Question

If $\int (\sqrt{x}+x\sqrt{x}+x)dx=\frac{x^a}{a}+\frac{x^b}{b}+\frac{x^c}{c}+K$ such that $x>0,$ then the value of $8abc$ is (where $K$ is constant of integration and $a,b$ and $c$ are fixed constants)

• A. 60.00
• B. 60.0
• C. 60

Answer 1

Answer: (A), (B), (C)

$\int (\sqrt{x}+x\sqrt{x}+x)dx=\int \sqrt{x}dx+\int x\sqrt{x}dx+\int xdx$
$=\frac{x^{3/2}}{\frac{3}{2}}+\frac{x^{3/2+1}}{\frac{3}{2}+1}+\frac{x^2}{2}+K$
$\Rightarrow \frac{x^a}{a}+\frac{x^b}{b}+\frac{x^c}{c}+K=\frac{x^{3/2}}{\frac{3}{2}}+\frac{x^{3/2+1}}{\frac{3}{2}+1}+\frac{x^2}{2}+K$
By comparing, we get
$a=\frac{3}{2},b=\frac{5}{2},c=2$
$\therefore 8abc=8\times \frac{3}{2}\times \frac{5}{2}\times 2=60$