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

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

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

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

- \frac{( x + 1 ) ( x - 1 )}{x} + 2 x + C

Solution steps

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

Apply Integration By Parts

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

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

\int - 2 d x = - 2 x

= - \frac{( x + 1 ) ( x - 1 )}{x} - (- 2 x)

Simplify

= - \frac{( x + 1 ) ( x - 1 )}{x} + 2 x

Add a constant to the solution

= - \frac{( x + 1 ) ( x - 1 )}{x} + 2 x + C

lo g_{9} (ln (x))

\frac{d}{d x} (1 - 4 x)

x \to + 0 lim (\frac{8}{6 + e ^{x}})

\int_{1}^{\infty} \frac{3}{x ^{4}} d x

\frac{( 17 - x ) 2 x + 8}{2}

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