What is the integral of 1+4/(3x)-1/(x^2) ?

The integral of 1+4/(3x)-1/(x^2) is x+4/3 ln|x|+1/x+C

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integral of 1+4/(3x)-1/(x^2)

\int 1 + \frac{4}{3 x} - \frac{1}{x ^{2}} d x

x + \frac{4}{3} ln ∣ x ∣ + \frac{1}{x} + C

Solution steps

\int 1 + \frac{4}{3 x} - \frac{1}{x ^{2}} 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 1 d x + \int \frac{4}{3 x} d x - \int \frac{1}{x ^{2}} d x

\int 1 d x = x

\int \frac{4}{3 x} d x = \frac{4}{3} ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= x + \frac{4}{3} ln ∣ x ∣ - (- \frac{1}{x})

Simplify

= x + \frac{4}{3} ln ∣ x ∣ + \frac{1}{x}

Add a constant to the solution

= x + \frac{4}{3} ln ∣ x ∣ + \frac{1}{x} + C

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What is the integral of 1+4/(3x)-1/(x^2) ?
The integral of 1+4/(3x)-1/(x^2) is x+4/3 ln|x|+1/x+C