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

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

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

\int \frac{x ^{2} + 4 x + 4}{x ^{2} + 4 x + 3} d x

x - \frac{1}{2} ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + 2 + C

Solution steps

\int \frac{x ^{2} + 4 x + 4}{x ^{2} + 4 x + 3} d x

Expand

\frac{x ^{2} + 4 x + 4}{x ^{2} + 4 x + 3} : \frac{x ^{2}}{x ^{2} + 4 x + 3} + \frac{4 x}{x ^{2} + 4 x + 3} + \frac{4}{x ^{2} + 4 x + 3}

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{x ^{2}}{x ^{2} + 4 x + 3} d x + \int \frac{4 x}{x ^{2} + 4 x + 3} d x + \int \frac{4}{x ^{2} + 4 x + 3} d x

\int \frac{x ^{2}}{x ^{2} + 4 x + 3} d x = x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)

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

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

= x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣) + 4 (\frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣) + 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣)

Simplify

x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣) + 4 (\frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣) + 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣) : x - \frac{1}{2} ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + 2

= x - \frac{1}{2} ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + 2

Add a constant to the solution

= x - \frac{1}{2} ln ∣ x + 3 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + 2 + C

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