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

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

Step 1 · Factor and set up partial fractions

$$ 2x^2+x-3 = (2x+3)(x-1), \qquad \frac{3x+2}{(2x+3)(x-1)} = \frac{A}{2x+3} + \frac{B}{x-1}. $$

Multiplying through: $3x+2=A(x-1)+B(2x+3)$. Setting $x=1$: $5=5B$, so $B=1$. Setting $x=-\tfrac32$: $-\tfrac52=-\tfrac52A$, so $A=1$.

Step 2 · Integrate

$$ I = \int \left(\frac{1}{2x+3} + \frac{1}{x-1}\right) dx = \frac12\ln|2x+3| + \ln|x-1| + C. $$

Step 3 · Combine logarithms

$$ I = \ln\left|(x-1)\sqrt{2x+3}\right| + C. $$

Summary Table

Additional Notes

• The domain excludes $x=-3/2$ and $x=1$, the roots of the denominator.
• Because the first partial-fraction term has a linear factor with leading coefficient $2$, its antiderivative picks up a factor of $\tfrac12$: $\int dx/(2x+3)=\tfrac12\ln|2x+3|$, not $\ln|2x+3|$ — a common place to drop a constant when the leading coefficient of a linear factor isn't $1$.
• The result can be checked by differentiating $\ln|x-1|+\tfrac12\ln|2x+3|$: this gives $\tfrac{1}{x-1}+\tfrac{1}{2x+3}=\tfrac{(2x+3)+(x-1)}{(x-1)(2x+3)}=\tfrac{3x+2}{(x-1)(2x+3)}$, matching the integrand since $(x-1)(2x+3)=2x^2+x-3$.