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Integrate \(4x^{3} + \frac 1 x\) with respect to \(x\)


Question

Integrate \(4x^{3} + \frac 1 x\) with respect to \(x\)

Options

A) \(\ln x\) + \(x^{4}\) + K

B) \(x^{-1}4\) + \(x^{4}\) + K

C) \(12x^{2}\) - \(x^{-2}\) + K

D) \(\frac 1 5x^{5}\) + \(x^{-2}\) + K

The correct answer is A.

Explanation:

\(f(x) = 4x^3 + \cfrac{1}{x}\)
\(\int \left( 4x^3 + \cfrac{1}{x} \right)dx = \cfrac{4x^{3+1}}{3+1} + \ln x + c\)
\(=\cfrac{4x^4}{4}+\ln x + c\)
\(=x^4+\ln x + C\)

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Dicussion (1)

  • \(f(x) = 4x^3 + \cfrac{1}{x}\)
    \(\int \left( 4x^3 + \cfrac{1}{x} \right)dx = \cfrac{4x^{3+1}}{3+1} + \ln x + c\)
    \(=\cfrac{4x^4}{4}+\ln x + c\)
    \(=x^4+\ln x + C\)

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