Evaluate the integral: I = integral (2x - 3)/(1 - x) dx
Evaluate the integral: $$ I = \int \frac{2x - 3}{1 - x} \, dx. $$
Explanation of Method
The same "match the denominator" strategy as Problem 35 applies here, but the denominator $1-x$ has a negative-slope linear term. Writing the numerator as a multiple of $1-x$ plus a remainder, $2x-3 = -2(1-x)+1$, isolates a constant term and a simple $\frac{1}{1-x}$ term.
- Rewrite the numerator as a multiple 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
First, we notice that the numerator $2x-3$ can be rewritten in terms of the denominator $1-x$. We express $2x-3$ as:
$$ 2x - 3 = -2(1-x) + 1. $$
This leads to:
$$ \frac{2x-3}{1-x} = \frac{-2(1-x)+1}{1-x} = -2 + \frac{1}{1-x}. $$
Thus, the integral becomes:
$$ I = \int \left(-2 + \frac{1}{1-x}\right) dx. $$
Step 2 · Integrate each term
Now, we integrate each term separately. The integral of $-2$ with respect to $x$ is:
$$ \int -2\,dx = -2x. $$
The integral of $\dfrac{1}{1-x}$ with respect to $x$ is:
$$ \int \frac{1}{1-x}\,dx = -\ln|1-x|. $$
Thus, the integral becomes:
$$ I = -2x - \ln|1-x| + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Antiderivative |
|---|---|
| $-2$ | $-2x$ |
| $1/(1-x)$ | $-\ln|1-x|$ |
Additional Notes
- The domain of $I$ requires $x \neq 1$, since the denominator vanishes there.
- The extra minus sign in $\int \frac{1}{1-x}\,dx=-\ln|1-x|$ (compared to $\int \frac{1}{x-1}\,dx=\ln|x-1|$) comes from the chain rule: the derivative of $1-x$ is $-1$, not $1$.
- The result can be checked by differentiating $-2x-\ln|1-x|$ and confirming it returns $\dfrac{2x-3}{1-x}$.
Result
The solution to the integral is: