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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int - 2 d m = - 2 m

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

Simplify

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

Add a constant to the solution

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

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