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

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

Step 1 · Rewrite the numerator

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

Step 2 · Split the integral

$$ I = \int \frac{2(x-2)}{(x-2)^2}\,dx + \int \frac{1}{(x-2)^2}\,dx = \int \frac{2}{x-2}\,dx + \int \frac{1}{(x-2)^2}\,dx. $$

Step 3 · Integrate each piece

$$ \int \frac{2}{x-2}\,dx = 2\ln|x-2|, \qquad \int \frac{1}{(x-2)^2}\,dx = -\frac{1}{x-2}. $$

Step 4 · Combine

$$ I = 2\ln|x-2| - \frac{1}{x-2} + C. $$

Summary Table

Additional Notes

• The domain excludes $x=2$, the repeated root of the denominator.
• This is the standard technique for a linear-over-repeated-linear-squared integrand: rewriting the numerator in terms of the shifted variable avoids full partial fraction machinery.
• The result can be checked by differentiating $2\ln|x-2|-1/(x-2)$: this gives $\tfrac{2}{x-2}+\tfrac{1}{(x-2)^2}=\tfrac{2(x-2)+1}{(x-2)^2}=\tfrac{2x-3}{(x-2)^2}$, matching the integrand.