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

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

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

integral

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

- \frac{1}{x} + 2 ln ∣ x ∣ + x + C

Solution steps

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

Expand

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 1 d x = x

= - \frac{1}{x} + 2 ln ∣ x ∣ + x

Add a constant to the solution

= - \frac{1}{x} + 2 ln ∣ x ∣ + x + C

\frac{( 1 - x ^{2} )}{( e ^{2} x )}

\frac{x ^{- 2}}{3}

sin (2 x) + cos (x)

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

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

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