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

The integral of ((x+4)(x-4))/(x^2) is ((x+4)(x-4))/x+(32)/x+C

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

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

\frac{( x + 4 ) ( x - 4 )}{x} + \frac{32}{x} + C

Solution steps

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

Apply Integration By Parts

: \frac{( x + 4 ) ( x - 4 )}{x} - \int \frac{32}{x ^{2}} d x

= \frac{( x + 4 ) ( x - 4 )}{x} - \int \frac{32}{x ^{2}} d x

\int \frac{32}{x ^{2}} d x = - \frac{32}{x}

= \frac{( x + 4 ) ( x - 4 )}{x} - (- \frac{32}{x})

Simplify

= \frac{( x + 4 ) ( x - 4 )}{x} + \frac{32}{x}

Add a constant to the solution

= \frac{( x + 4 ) ( x - 4 )}{x} + \frac{32}{x} + C

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What is the integral of ((x+4)(x-4))/(x^2) ?
The integral of ((x+4)(x-4))/(x^2) is ((x+4)(x-4))/x+(32)/x+C