What is the integral of (3x-1)/((x+4)^2) ?

The integral of (3x-1)/((x+4)^2) is 3ln|x+4|+(13)/(x+4)+C

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integral of (3x-1)/((x+4)^2)

\int \frac{3 x - 1}{( x + 4 ) ^{2}} d x

3 ln ∣ x + 4 ∣ + \frac{13}{x + 4} + C

Solution steps

\int \frac{3 x - 1}{( x + 4 ) ^{2}} d x

Expand

\frac{3 x - 1}{( x + 4 ) ^{2}} : \frac{3 x}{( x + 4 ) ^{2}} - \frac{1}{( x + 4 ) ^{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{3 x}{( x + 4 ) ^{2}} d x - \int \frac{1}{( x + 4 ) ^{2}} d x

\int \frac{3 x}{( x + 4 ) ^{2}} d x = 3 (ln ∣ x + 4 ∣ + \frac{4}{x + 4})

\int \frac{1}{( x + 4 ) ^{2}} d x = - \frac{1}{x + 4}

= 3 (ln ∣ x + 4 ∣ + \frac{4}{x + 4}) - (- \frac{1}{x + 4})

Simplify

3 (ln ∣ x + 4 ∣ + \frac{4}{x + 4}) - (- \frac{1}{x + 4}) : 3 ln ∣ x + 4 ∣ + \frac{13}{x + 4}

= 3 ln ∣ x + 4 ∣ + \frac{13}{x + 4}

Add a constant to the solution

= 3 ln ∣ x + 4 ∣ + \frac{13}{x + 4} + C

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What is the integral of (3x-1)/((x+4)^2) ?
The integral of (3x-1)/((x+4)^2) is 3ln|x+4|+(13)/(x+4)+C