What is the integral of (3k+3)/(k^2+8k+16) ?

The integral of (3k+3)/(k^2+8k+16) is 3ln|k+4|+9/(k+4)+C

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integral of (3k+3)/(k^2+8k+16)

\int \frac{3 k + 3}{k ^{2} + 8 k + 16} d k

3 ln ∣ k + 4 ∣ + \frac{9}{k + 4} + C

Solution steps

\int \frac{3 k + 3}{k ^{2} + 8 k + 16} d k

Expand

\frac{3 k + 3}{k ^{2} + 8 k + 16} : \frac{3 k}{k ^{2} + 8 k + 16} + \frac{3}{k ^{2} + 8 k + 16}

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 k}{k ^{2} + 8 k + 16} d k + \int \frac{3}{k ^{2} + 8 k + 16} d k

\int \frac{3 k}{k ^{2} + 8 k + 16} d k = 3 (ln ∣ k + 4 ∣ + \frac{4}{k + 4})

\int \frac{3}{k ^{2} + 8 k + 16} d k = - \frac{3}{k + 4}

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

Simplify

3 (ln ∣ k + 4 ∣ + \frac{4}{k + 4}) - \frac{3}{k + 4} : 3 ln ∣ k + 4 ∣ + \frac{9}{k + 4}

= 3 ln ∣ k + 4 ∣ + \frac{9}{k + 4}

Add a constant to the solution

= 3 ln ∣ k + 4 ∣ + \frac{9}{k + 4} + C

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What is the integral of (3k+3)/(k^2+8k+16) ?
The integral of (3k+3)/(k^2+8k+16) is 3ln|k+4|+9/(k+4)+C