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

The integral of (3z+1)/(z^2+4z+4) is 3ln|z+2|+5/(z+2)+C

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integral of (3z+1)/(z^2+4z+4)

\int \frac{3 z + 1}{z ^{2} + 4 z + 4} d z

3 ln ∣ z + 2 ∣ + \frac{5}{z + 2} + C

Solution steps

\int \frac{3 z + 1}{z ^{2} + 4 z + 4} d z

Expand

\frac{3 z + 1}{z ^{2} + 4 z + 4} : \frac{3 z}{z ^{2} + 4 z + 4} + \frac{1}{z ^{2} + 4 z + 4}

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 z}{z ^{2} + 4 z + 4} d z + \int \frac{1}{z ^{2} + 4 z + 4} d z

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

\int \frac{1}{z ^{2} + 4 z + 4} d z = - \frac{1}{z + 2}

= 3 (ln ∣ z + 2 ∣ + \frac{2}{z + 2}) - \frac{1}{z + 2}

Simplify

3 (ln ∣ z + 2 ∣ + \frac{2}{z + 2}) - \frac{1}{z + 2} : 3 ln ∣ z + 2 ∣ + \frac{5}{z + 2}

= 3 ln ∣ z + 2 ∣ + \frac{5}{z + 2}

Add a constant to the solution

= 3 ln ∣ z + 2 ∣ + \frac{5}{z + 2} + C

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What is the integral of (3z+1)/(z^2+4z+4) ?
The integral of (3z+1)/(z^2+4z+4) is 3ln|z+2|+5/(z+2)+C