\(\displaystyle\int \frac{2x^2 - 3x + 5}{\sqrt{x}}\,dx,\ x \gt 0\), is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{4}{5}x^{5/2} - 2x^{3/2} + 10\sqrt{x} + C\)
Dividing each term of the numerator by \(\sqrt{x} = x^{1/2}\):
\( \displaystyle \frac{2x^2 - 3x + 5}{\sqrt{x}} = 2x^{3/2} - 3x^{1/2} + 5x^{-1/2} \)
By the power rule \(\int x^{n}\,dx = \frac{x^{n+1}}{n+1}\), term by term:
\( \displaystyle \int 2x^{3/2}\,dx = 2 \cdot \frac{x^{5/2}}{5/2} = \frac{4}{5}x^{5/2} \)
\( \displaystyle \int 3x^{1/2}\,dx = 3 \cdot \frac{x^{3/2}}{3/2} = 2x^{3/2} \)
\( \displaystyle \int 5x^{-1/2}\,dx = 5 \cdot \frac{x^{1/2}}{1/2} = 10\sqrt{x} \)
So
\( \displaystyle \int \frac{2x^2 - 3x + 5}{\sqrt{x}}\,dx = \frac{4}{5}x^{5/2} - 2x^{3/2} + 10\sqrt{x} + C \)