What is the integral of (p^{-1}+4/x) ?

The integral of (p^{-1}+4/x) is x/p+4ln|x|+C

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integral of (p^{-1}+4/x)

\int (p^{- 1} + \frac{4}{x}) d x

\frac{x}{p} + 4 ln ∣ x ∣ + C

Solution steps

\int p^{- 1} + \frac{4}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int p^{- 1} d x + \int \frac{4}{x} d x

\int p^{- 1} d x = \frac{x}{p}

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= \frac{x}{p} + 4 ln ∣ x ∣

Add a constant to the solution

= \frac{x}{p} + 4 ln ∣ x ∣ + C

\int x \cdot x + 2 + 2 \cdot x + 2 d x

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\int x 64 - x^{2} d x

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What is the integral of (p^{-1}+4/x) ?
The integral of (p^{-1}+4/x) is x/p+4ln|x|+C