Evaluate the integral: I = integral dx/(3x²-2x-1)

Evaluate the integral: $$ I = \int \frac{dx}{3x^2-2x-1}. $$

Step 1 · Factor by grouping

$$ 3x^2-2x-1 = 3x^2-3x+x-1 = 3x(x-1)+1(x-1) = (3x+1)(x-1). $$

Step 2 · Partial fraction decomposition

Assume $\dfrac{1}{(3x+1)(x-1)}=\dfrac{A}{3x+1}+\dfrac{B}{x-1}$. Multiplying through:

$$ 1 = A(x-1) + B(3x+1) = (A+3B)x + (-A+B). $$

Equating coefficients: $A+3B=0$ and $-A+B=1$. Solving, $A=-3B$, so $4B=1$, giving $B=\tfrac14$ and $A=-\tfrac34$.

Step 3 · Integrate each term

$$ I = \int \frac{-3/4}{3x+1}\,dx + \int \frac{1/4}{x-1}\,dx = -\frac14\ln|3x+1| + \frac14\ln|x-1| + C. $$

Step 4 · Combine logarithms

$$ I = \frac14\ln\left|\frac{x-1}{3x+1}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=-1/3$ and $x=1$, the roots of the denominator.
• Factoring by grouping ($3x^2-3x+x-1$) is a reliable alternative to the quadratic formula whenever the quadratic has rational roots that aren't immediately obvious.
• The result can be checked by differentiating $\tfrac14\ln|x-1|-\tfrac14\ln|3x+1|$: this gives $\tfrac{1}{4(x-1)}-\tfrac{3}{4(3x+1)}=\tfrac{(3x+1)-3(x-1)}{4(x-1)(3x+1)}=\tfrac{4}{4(x-1)(3x+1)}=\tfrac{1}{(x-1)(3x+1)}$, matching the integrand since $(x-1)(3x+1)=3x^2-2x-1$.