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

The integral of ((t+2)(t-4))/(t^2) is -((t+2)(t-4))/t+2t-2ln|t|+C

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

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

- \frac{( t + 2 ) ( t - 4 )}{t} + 2 t - 2 ln ∣ t ∣ + C

Solution steps

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

Apply Integration By Parts

: - \frac{( t + 2 ) ( t - 4 )}{t} - \int - \frac{2 t - 2}{t} d t

= - \frac{( t + 2 ) ( t - 4 )}{t} - \int - \frac{2 t - 2}{t} d t

\int - \frac{2 t - 2}{t} d t = - 2 t + 2 ln ∣ t ∣

= - \frac{( t + 2 ) ( t - 4 )}{t} - (- 2 t + 2 ln ∣ t ∣)

Simplify

= - \frac{( t + 2 ) ( t - 4 )}{t} + 2 t - 2 ln ∣ t ∣

Add a constant to the solution

= - \frac{( t + 2 ) ( t - 4 )}{t} + 2 t - 2 ln ∣ t ∣ + C

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What is the integral of ((t+2)(t-4))/(t^2) ?
The integral of ((t+2)(t-4))/(t^2) is -((t+2)(t-4))/t+2t-2ln|t|+C