What is the integral of (x^2+6x+5)/(x^2+2x+1) ?

The integral of (x^2+6x+5)/(x^2+2x+1) is -(x^2)/(x+1)+2x+2+4ln|x+1|+1/(x+1)+C

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integral of (x^2+6x+5)/(x^2+2x+1)

\int \frac{x ^{2} + 6 x + 5}{x ^{2} + 2 x + 1} d x

- \frac{x ^{2}}{x + 1} + 2 x + 2 + 4 ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

Solution steps

\int \frac{x ^{2} + 6 x + 5}{x ^{2} + 2 x + 1} d x

Expand

\frac{x ^{2} + 6 x + 5}{x ^{2} + 2 x + 1} : \frac{x ^{2}}{x ^{2} + 2 x + 1} + \frac{6 x}{x ^{2} + 2 x + 1} + \frac{5}{x ^{2} + 2 x + 1}

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} + 2 x + 1} d x + \int \frac{6 x}{x ^{2} + 2 x + 1} d x + \int \frac{5}{x ^{2} + 2 x + 1} d x

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

\int \frac{6 x}{x ^{2} + 2 x + 1} d x = 6 (ln ∣ x + 1 ∣ + \frac{1}{x + 1})

\int \frac{5}{x ^{2} + 2 x + 1} d x = - \frac{5}{x + 1}

= - \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣) + 6 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{5}{x + 1}

Simplify

- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣) + 6 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{5}{x + 1} : - \frac{x ^{2}}{x + 1} + 2 x + 2 + 4 ln ∣ x + 1 ∣ + \frac{1}{x + 1}

= - \frac{x ^{2}}{x + 1} + 2 x + 2 + 4 ln ∣ x + 1 ∣ + \frac{1}{x + 1}

Add a constant to the solution

= - \frac{x ^{2}}{x + 1} + 2 x + 2 + 4 ln ∣ x + 1 ∣ + \frac{1}{x + 1} + C

What is the integral of (x^2+6x+5)/(x^2+2x+1) ?
The integral of (x^2+6x+5)/(x^2+2x+1) is -(x^2)/(x+1)+2x+2+4ln|x+1|+1/(x+1)+C