What is the integral of ((t+9)^2)/(t^2) ?

The integral of ((t+9)^2)/(t^2) is t+18ln|t|-(81)/t+C

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integral of ((t+9)^2)/(t^2)

\int \frac{( t + 9 ) ^{2}}{t ^{2}} d t

t + 18 ln ∣ t ∣ - \frac{81}{t} + C

Solution steps

\int \frac{( t + 9 ) ^{2}}{t ^{2}} d t

Expand

\frac{( t + 9 ) ^{2}}{t ^{2}} : 1 + \frac{18}{t} + \frac{81}{t ^{2}}

= \int 1 + \frac{18}{t} + \frac{81}{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 1 d t + \int \frac{18}{t} d t + \int \frac{81}{t ^{2}} d t

\int 1 d t = t

\int \frac{18}{t} d t = 18 ln ∣ t ∣

\int \frac{81}{t ^{2}} d t = - \frac{81}{t}

= t + 18 ln ∣ t ∣ - \frac{81}{t}

Add a constant to the solution

= t + 18 ln ∣ t ∣ - \frac{81}{t} + C

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What is the integral of ((t+9)^2)/(t^2) ?
The integral of ((t+9)^2)/(t^2) is t+18ln|t|-(81)/t+C