Question

$\int \frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}dx$ is equal to

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

Answer 1

The correct option is A $\frac{x^2}{2}-x+C$

$\int \frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}dx=\int \frac{(\sqrt{x}+1)\sqrt{x}\left(\right(\sqrt{x}{)}^3-1)}{\sqrt{x}(x+\sqrt{x}+1)}dx=\int \frac{(\sqrt{x}+1)(\sqrt{x}-1)(x+1+\sqrt{x})}{(x+\sqrt{x}+1)}dx=\int (x-1)dx=\frac{x^2}{2}-x+C$