Integrate \(\frac{x^2 -\sqrt{x}}{x}\) with respect to x
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Correct Answer: Option A
Explanation:
\(\int \frac{x^2 -\sqrt{x}}{x} = \int \frac{x^2}{x} - \frac{x^{\frac{1}{2}}}{x}\\
\int x - x^{\frac{-1}{2}}\\
=\left(\frac{1}{2}\right)x^2 - \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+K\\
=\frac{x^2}{2}-2x^{\frac{1}{2}}+K\\
=\frac{x^2}{2}-2\sqrt{x}+K\)
\(\int \frac{x^2 -\sqrt{x}}{x} = \int \frac{x^2}{x} - \frac{x^{\frac{1}{2}}}{x}\\
\int x - x^{\frac{-1}{2}}\\
=\left(\frac{1}{2}\right)x^2 - \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+K\\
=\frac{x^2}{2}-2x^{\frac{1}{2}}+K\\
=\frac{x^2}{2}-2\sqrt{x}+K\)