What is the integral of (5x-3)/(x^2+6x+5) ?

The integral of (5x-3)/(x^2+6x+5) is -2ln|x+1|+7ln|x+5|+C

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integral of (5x-3)/(x^2+6x+5)

\int \frac{5 x - 3}{x ^{2} + 6 x + 5} d x

- 2 ln ∣ x + 1 ∣ + 7 ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{5 x - 3}{x ^{2} + 6 x + 5} d x

Expand

\frac{5 x - 3}{x ^{2} + 6 x + 5} : \frac{5 x}{x ^{2} + 6 x + 5} - \frac{3}{x ^{2} + 6 x + 5}

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{5 x}{x ^{2} + 6 x + 5} d x - \int \frac{3}{x ^{2} + 6 x + 5} d x

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

\int \frac{3}{x ^{2} + 6 x + 5} d x = 3 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣)

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

Simplify

5 (\frac{5}{4} ln ∣ x + 5 ∣ - \frac{1}{4} ln ∣ x + 1 ∣) - 3 (\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣) : - 2 ln ∣ x + 1 ∣ + 7 ln ∣ x + 5 ∣

= - 2 ln ∣ x + 1 ∣ + 7 ln ∣ x + 5 ∣

Add a constant to the solution

= - 2 ln ∣ x + 1 ∣ + 7 ln ∣ x + 5 ∣ + C

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What is the integral of (5x-3)/(x^2+6x+5) ?
The integral of (5x-3)/(x^2+6x+5) is -2ln|x+1|+7ln|x+5|+C