What is the integral of (-4n-2)/((n-3)^2) ?

The integral of (-4n-2)/((n-3)^2) is -4ln|n-3|+(14)/(n-3)+C

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integral of (-4n-2)/((n-3)^2)

\int \frac{- 4 n - 2}{( n - 3 ) ^{2}} d n

- 4 ln ∣ n - 3 ∣ + \frac{14}{n - 3} + C

Solution steps

\int \frac{- 4 n - 2}{( n - 3 ) ^{2}} d n

Expand

\frac{- 4 n - 2}{( n - 3 ) ^{2}} : - \frac{4 n}{( n - 3 ) ^{2}} - \frac{2}{( n - 3 ) ^{2}}

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{4 n}{( n - 3 ) ^{2}} d n - \int \frac{2}{( n - 3 ) ^{2}} d n

\int \frac{4 n}{( n - 3 ) ^{2}} d n = 4 (ln ∣ n - 3 ∣ - \frac{3}{n - 3})

\int \frac{2}{( n - 3 ) ^{2}} d n = - \frac{2}{n - 3}

= - 4 (ln ∣ n - 3 ∣ - \frac{3}{n - 3}) - (- \frac{2}{n - 3})

Simplify

- 4 (ln ∣ n - 3 ∣ - \frac{3}{n - 3}) - (- \frac{2}{n - 3}) : - 4 ln ∣ n - 3 ∣ + \frac{14}{n - 3}

= - 4 ln ∣ n - 3 ∣ + \frac{14}{n - 3}

Add a constant to the solution

= - 4 ln ∣ n - 3 ∣ + \frac{14}{n - 3} + C

What is the integral of (-4n-2)/((n-3)^2) ?
The integral of (-4n-2)/((n-3)^2) is -4ln|n-3|+(14)/(n-3)+C