What is the integral of ((3+r)/r)^2 ?

The integral of ((3+r)/r)^2 is -9/r+6ln|r|+r+C

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integral of ((3+r)/r)^2

\int (\frac{3 + r}{r})^{2} d r

- \frac{9}{r} + 6 ln ∣ r ∣ + r + C

Solution steps

\int (\frac{3 + r}{r})^{2} d r

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int \frac{( 3 + r ) ^{2}}{r ^{2}} d r

Expand

\frac{( 3 + r ) ^{2}}{r ^{2}} : \frac{9}{r ^{2}} + \frac{6}{r} + 1

= \int \frac{9}{r ^{2}} + \frac{6}{r} + 1 d r

Apply the Sum Rule:

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

= \int \frac{9}{r ^{2}} d r + \int \frac{6}{r} d r + \int 1 d r

\int \frac{9}{r ^{2}} d r = - \frac{9}{r}

\int \frac{6}{r} d r = 6 ln ∣ r ∣

\int 1 d r = r

= - \frac{9}{r} + 6 ln ∣ r ∣ + r

Add a constant to the solution

= - \frac{9}{r} + 6 ln ∣ r ∣ + r + C

What is the integral of ((3+r)/r)^2 ?
The integral of ((3+r)/r)^2 is -9/r+6ln|r|+r+C