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

The integral of (t-2/t)(t+2/t) is (t^3)/3+4/t+C

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

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

\frac{t ^{3}}{3} + \frac{4}{t} + C

Solution steps

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

Expand

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

= \int t^{2} - \frac{4}{t ^{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int t^{2} d t - \int \frac{4}{t ^{2}} d t

\int t^{2} d t = \frac{t ^{3}}{3}

\int \frac{4}{t ^{2}} d t = - \frac{4}{t}

= \frac{t ^{3}}{3} - (- \frac{4}{t})

Simplify

= \frac{t ^{3}}{3} + \frac{4}{t}

Add a constant to the solution

= \frac{t ^{3}}{3} + \frac{4}{t} + C

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What is the integral of (t-2/t)(t+2/t) ?
The integral of (t-2/t)(t+2/t) is (t^3)/3+4/t+C