Integrate {x} - \; x/2 + \; {2}{{x}}

Correct Answer is : 23xx− 14x2+4x+c\; 2/3 x {x} - \; 1/4 x² + 4 {x} + c32xx−41x2+4x+c

\; 2/3 x {x} - \; 1/4 x² + 4 {x} + c

\; 3/2 {x} - \; 1/4 x² + 4 {x} + c

\; 2/3 x {x} - \; 1/2 x² - \; 1/2 {x} + c

Correct Answer is : 23xx− 14x2+4x+c\; 2/3 x {x} - \; 1/4 x² + 4 {x} + c32xx−41x2+4x+c

Test Index

Integral Calculus

Question : 51 of 73

Marks: +1, -0

Integrate

\sqrt{x} - \; \frac{x}{2} + \; \frac{2}{\sqrt{x}}

$\; \frac{2}{3} x \sqrt{x} - \; \frac{1}{4} x^{2} + 4 \sqrt{x} + c$
$\; \frac{3}{2} \sqrt{x} - \; \frac{1}{4} x^{2} + 4 \sqrt{x} + c$
$\; \frac{2}{3} x \sqrt{x} - \; \frac{1}{2} x^{2} - \; \frac{1}{2} \sqrt{x} + c$
None

Solution:

Concept:

Integrate each term separately using power rule:

\int x^n dx = \frac{x^{n+1}}{n+1} + c

Explanation:

\int \left( \sqrt{x} - \frac{x}{2} + \frac{2}{\sqrt{x}} \right) dx

Rewrite terms as powers:

\sqrt{x} = x^{1/2}

\frac{x}{2} = \frac{1}{2}x

\frac{2}{\sqrt{x}} = 2 x^{-1/2}

Integrate each:

\int x^{1/2} dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} = \frac{2}{3}x\sqrt{x}

\int \frac{1}{2}x dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}

\int 2x^{-1/2} dx = 2 \cdot \frac{x^{1/2}}{1/2} = 2 \cdot 2 x^{1/2} = 4\sqrt{x}

Combine results with constant of integration

c

\frac{2}{3}x\sqrt{x} - \frac{1}{4}x^2 + 4\sqrt{x} + c

Answer:

\frac{2}{3}x\sqrt{x} - \frac{1}{4}x^2 + 4\sqrt{x} + c

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