What is the integral of (8x^2-1)/(x^2+4x+3) ?

The integral of (8x^2-1)/(x^2+4x+3) is 8x+39/2 ln|x+1|-39/2 ln|x+3|-16ln|x^2+4x+3|+16+C

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integral of (8x^2-1)/(x^2+4x+3)

\int \frac{8 x ^{2} - 1}{x ^{2} + 4 x + 3} d x

8 x + \frac{39}{2} ln ∣ x + 1 ∣ - \frac{39}{2} ln ∣ x + 3 ∣ - 16 ln x^{2} + 4 x + 3 + 16 + C

Solution steps

\int \frac{8 x ^{2} - 1}{x ^{2} + 4 x + 3} d x

Expand

\frac{8 x ^{2} - 1}{x ^{2} + 4 x + 3} : \frac{8 x ^{2}}{x ^{2} + 4 x + 3} - \frac{1}{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{8 x ^{2}}{x ^{2} + 4 x + 3} d x - \int \frac{1}{x ^{2} + 4 x + 3} d x

\int \frac{8 x ^{2}}{x ^{2} + 4 x + 3} d x = 8 (x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣))

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

= 8 (x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)) - (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣)

Simplify

8 (x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)) - (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣) : 8 x + \frac{39}{2} ln ∣ x + 1 ∣ - \frac{39}{2} ln ∣ x + 3 ∣ - 16 ln x^{2} + 4 x + 3 + 16

= 8 x + \frac{39}{2} ln ∣ x + 1 ∣ - \frac{39}{2} ln ∣ x + 3 ∣ - 16 ln x^{2} + 4 x + 3 + 16

Add a constant to the solution

= 8 x + \frac{39}{2} ln ∣ x + 1 ∣ - \frac{39}{2} ln ∣ x + 3 ∣ - 16 ln x^{2} + 4 x + 3 + 16 + C

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What is the integral of (8x^2-1)/(x^2+4x+3) ?
The integral of (8x^2-1)/(x^2+4x+3) is 8x+39/2 ln|x+1|-39/2 ln|x+3|-16ln|x^2+4x+3|+16+C