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

The integral of (-3x-1)/(2x^2+3x+1) is -3/4 ln|16x^2+24x+8|-5/4 ln|4x+4|+5/4 ln|4x+2|+C

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

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

- \frac{3}{4} ln 16 x^{2} + 24 x + 8 - \frac{5}{4} ln ∣ 4 x + 4 ∣ + \frac{5}{4} ln ∣ 4 x + 2 ∣ + C

Solution steps

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

Expand

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

\int \frac{3 x}{2 x ^{2} + 3 x + 1} d x = 6 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣))

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

= - 6 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣)) - (- ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣)

Simplify

- 6 (\frac{1}{8} ln 16 x^{2} + 24 x + 8 + \frac{3}{8} (ln ∣ 4 x + 4 ∣ - ln ∣ 4 x + 2 ∣)) - (- ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣) : - \frac{3}{4} ln 16 x^{2} + 24 x + 8 - \frac{5}{4} ln ∣ 4 x + 4 ∣ + \frac{5}{4} ln ∣ 4 x + 2 ∣

= - \frac{3}{4} ln 16 x^{2} + 24 x + 8 - \frac{5}{4} ln ∣ 4 x + 4 ∣ + \frac{5}{4} ln ∣ 4 x + 2 ∣

Add a constant to the solution

= - \frac{3}{4} ln 16 x^{2} + 24 x + 8 - \frac{5}{4} ln ∣ 4 x + 4 ∣ + \frac{5}{4} ln ∣ 4 x + 2 ∣ + C

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What is the integral of (-3x-1)/(2x^2+3x+1) ?
The integral of (-3x-1)/(2x^2+3x+1) is -3/4 ln|16x^2+24x+8|-5/4 ln|4x+4|+5/4 ln|4x+2|+C