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

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

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

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

x + \frac{4}{x} + ln ∣ x ∣ + \frac{2}{x ^{2}} + C

Solution steps

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

Expand

(1 + \frac{1}{x}) (1 - \frac{4}{x ^{2}}) : 1 - \frac{4}{x ^{2}} + \frac{1}{x} - \frac{4}{x ^{3}}

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

\int 1 d x = x

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

\int \frac{1}{x} d x = ln ∣ x ∣

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

= x - (- \frac{4}{x}) + ln ∣ x ∣ - (- \frac{2}{x ^{2}})

Simplify

= x + \frac{4}{x} + ln ∣ x ∣ + \frac{2}{x ^{2}}

Add a constant to the solution

= x + \frac{4}{x} + ln ∣ x ∣ + \frac{2}{x ^{2}} + C

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