Evaluate the integral: I = integral (x - 2)/(x + 3) dx
Evaluate the integral: $$ I = \int \frac{x - 2}{x + 3} \, dx. $$
Explanation of Method
An improper rational function with linear numerator and denominator is simplified by writing the numerator as the denominator plus a remainder: $x-2 = (x+3)-5$. Dividing through gives a constant term plus a simple $\frac{1}{x+3}$ term, which integrates to a logarithm.
- Rewrite the numerator in terms of the denominator plus a remainder.
- Split the fraction into a constant term and a proper fraction.
- Integrate each term separately.
- Combine the results and add the constant of integration.
Step 1 · Simplify the integrand
We begin by simplifying the expression inside the integral. We can rewrite the fraction as:
$$ \frac{x-2}{x+3} = \frac{(x+3)-5}{x+3} = \frac{x+3}{x+3} - \frac{5}{x+3}. $$
Thus, the integral becomes:
$$ I = \int \left(1 - \frac{5}{x+3}\right) dx. $$
Step 2 · Integrate each term
Now we can integrate each term separately. The integral of $1$ with respect to $x$ is:
$$ \int 1\,dx = x. $$
The integral of $\dfrac{5}{x+3}$ with respect to $x$ is:
$$ \int \frac{5}{x+3}\,dx = 5\ln|x+3|. $$
Thus, the integral becomes:
$$ I = x - 5\ln|x+3| + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Antiderivative |
|---|---|
| $1$ | $x$ |
| $-5/(x+3)$ | $-5\ln|x+3|$ |
Additional Notes
- The domain of $I$ requires $x \neq -3$, since the denominator vanishes there.
- This "add and subtract" strategy is the linear-denominator analogue of the improper-fraction tricks used for quadratic denominators in Problems 6, 7, 9, and 10.
- The result can be checked by differentiating $x-5\ln|x+3|$ and confirming it returns $\dfrac{x-2}{x+3}$.
Result
The solution to the integral is: