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

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

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

integral

\frac{( 1 + x )}{x ^{2}}

- \frac{1}{x} + ln ∣ x ∣ + C

Solution steps

\int \frac{1 + x}{x ^{2}} d x

Expand

\frac{1 + x}{x ^{2}} : \frac{1}{x ^{2}} + \frac{1}{x}

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

= - \frac{1}{x} + ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{x} + ln ∣ x ∣ + C

20 - 2 x

\frac{1}{2 x ^{2}}

20 (x - 3) (x + 2)^{3}

\frac{( 3 x ^{2} - 2 )}{( x ^{2} - 2 x - 8 )}

\frac{1}{( tan ( x ) + cot ( x ) ) ^{2}}

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