Integration of {1}{{x}}(x²-3x+[3]{x}+7) is

{{x}}{2} ≤ft( {x³}{3} - {3x²}{2} + 3/4 x^{4/3} + 7x )

{x³}{3} - 2x² + 3x¹/3} + 7x + k

{2 x^{5/2}}{5} - 2 x³/2} + {6 x^{5/6}}{5} + 14 x¹/2} + k

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Integral Calculus

Question : 7 of 73

Marks: +1, -0

Integration of

\frac{1}{\sqrt{x}}(x^{2}-3x+\sqrt[3]{x}+7)

$\frac{\sqrt{x}}{2} \left( \frac{x^{3}}{3} - \frac{3x^{2}}{2} + \frac{3}{4} x^{4/3} + 7x \right)$
$\frac{x^{3}}{3} - 2x^{2} + 3x^{1/3} + 7x + k$
$\frac{2 x^{5/2}}{5} - 2 x^{3/2} + \frac{6 x^{5/6}}{5} + 14 x^{1/2} + k$
None

Solution:

Concept:

Integrate the given expression by rewriting each term in power form and applying the power rule.

Explanation:

First, rewrite

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

and multiply through:

x^{-1/2} \cdot x^{2} = x^{3/2}

x^{-1/2} \cdot (-3x) = -3x^{1/2}

x^{-1/2} \cdot \sqrt[3]{x} = x^{-1/6}

x^{-1/2} \cdot 7 = 7x^{-1/2}

So the integrand is

x^{3/2} - 3x^{1/2} + x^{-1/6} + 7x^{-1/2}

Integrate term by term:

\int x^{3/2} dx = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}

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

\int x^{-1/6} dx = \frac{x^{5/6}}{5/6} = \frac{6}{5}x^{5/6}

\int 7x^{-1/2} dx = 7 \cdot \frac{x^{1/2}}{1/2} = 14x^{1/2}

Add constant

k

\frac{2}{5}x^{5/2} - 2x^{3/2} + \frac{6}{5}x^{5/6} + 14x^{1/2} + k

Answer:

Option C matches exactly.

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