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

The integral of (t+1/t)^2 is (t^3)/3+2t-1/t+C

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integral of (t+1/t)^2

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

\frac{t ^{3}}{3} + 2 t - \frac{1}{t} + C

Solution steps

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

Expand

(t + \frac{1}{t})^{2} : t^{2} + 2 + \frac{1}{t ^{2}}

= \int t^{2} + 2 + \frac{1}{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 2 d t + \int \frac{1}{t ^{2}} d t

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

\int 2 d t = 2 t

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

= \frac{t ^{3}}{3} + 2 t - \frac{1}{t}

Add a constant to the solution

= \frac{t ^{3}}{3} + 2 t - \frac{1}{t} + C

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What is the integral of (t+1/t)^2 ?
The integral of (t+1/t)^2 is (t^3)/3+2t-1/t+C