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

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

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

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

- \frac{5}{6} ln ∣ 3 x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{3} + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{1}{4} (- ln ∣ 3 x + 1 ∣ + ln ∣ x - 1 ∣) d x = - \frac{1}{12} (3 x ln ∣ x - 1 ∣ - 3 x ln ∣ 3 x + 1 ∣ - ln ∣ 3 x + 1 ∣ - 3 ln ∣ x - 1 ∣ + 4)

= (3 - x) (- \frac{1}{4} ln ∣ 3 x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) - (- \frac{1}{12} (3 x ln ∣ x - 1 ∣ - 3 x ln ∣ 3 x + 1 ∣ - ln ∣ 3 x + 1 ∣ - 3 ln ∣ x - 1 ∣ + 4))

Simplify

(3 - x) (- \frac{1}{4} ln ∣ 3 x + 1 ∣ + \frac{1}{4} ln ∣ x - 1 ∣) - (- \frac{1}{12} (3 x ln ∣ x - 1 ∣ - 3 x ln ∣ 3 x + 1 ∣ - ln ∣ 3 x + 1 ∣ - 3 ln ∣ x - 1 ∣ + 4)) : - \frac{5}{6} ln ∣ 3 x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{3}

= - \frac{5}{6} ln ∣ 3 x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{3}

Add a constant to the solution

= - \frac{5}{6} ln ∣ 3 x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{3} + C

x \to 1 lim ((2 x + 1) (x - 1))

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

\frac{1}{3} x^{3} - \frac{7}{2} x^{2} - 8 x + 3

\int e^{1 - 6 t} d t

(x^{3} - 7 x + 2) (2 x^{4} - 1)

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