Evaluate the integral: I = integral (2x+7)/(x²+x-2) dx

Evaluate the integral: $$ I = \int \frac{2x+7}{x^2+x-2}\,dx. $$

Step 1 · Factor and set up partial fractions

$$ x^2+x-2 = (x-1)(x+2), \qquad \frac{2x+7}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}. $$

Multiplying through: $2x+7=A(x+2)+B(x-1)$. Setting $x=1$: $9=3A$, so $A=3$. Setting $x=-2$: $3=-3B$, so $B=-1$.

Step 2 · Integrate

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

Step 3 · Combine logarithms

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

Summary Table

Additional Notes

• The domain excludes $x=1$ and $x=-2$, the roots of the denominator.
• Since the numerator's degree is lower than the denominator's, no polynomial long division is needed before the partial fraction decomposition — a direct advantage over problems like $\int(x^2+x)\ln(x+1)\,dx$-style setups where division is unavoidable.
• The result can be checked by differentiating $3\ln|x-1|-\ln|x+2|$: this gives $\tfrac{3}{x-1}-\tfrac1{x+2}=\tfrac{3(x+2)-(x-1)}{(x-1)(x+2)}=\tfrac{2x+7}{(x-1)(x+2)}$, matching the integrand.