Evaluate the integral: I = integral dx/((x-1)(x+3))
Evaluate the integral: $$ I = \int \frac{dx}{(x-1)(x+3)}. $$
Explanation of Method
Since the denominator is a product of two distinct linear factors, partial fraction decomposition splits the integrand into two simple $1/(x-a)$ terms, each of which integrates to a logarithm.
- Set up the partial fraction decomposition.
- Solve for the unknown coefficients.
- Integrate each term.
- Combine using logarithm properties.
Step 1 · Set up the decomposition
We express the fraction as a sum of simpler fractions:
$$ \frac{1}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3}. $$
Multiplying both sides by $(x-1)(x+3)$:
$$ 1 = A(x+3) + B(x-1). $$
Step 2 · Solve for the coefficients
Setting $x=1$: $\;1=4A \Rightarrow A=\tfrac14$. Setting $x=-3$: $\;1=-4B \Rightarrow B=-\tfrac14$.
Step 3 · Rewrite and integrate
Substituting the coefficients:
$$ I = \frac14\int \frac{dx}{x-1} - \frac14\int \frac{dx}{x+3} = \frac14\ln|x-1| - \frac14\ln|x+3| + C. $$
Step 4 · Combine using logarithm properties
Combining the two logarithms:
$$ I = \frac14\ln\left|\frac{x-1}{x+3}\right| + C. $$
Summary Table
| Coefficient | Value |
|---|---|
| $A$ | $1/4$ |
| $B$ | $-1/4$ |
| Antiderivative | $\tfrac14\ln\left|\tfrac{x-1}{x+3}\right|$ |
Additional Notes
- The domain excludes $x=1$ and $x=-3$, where the original denominator vanishes.
- Whenever a rational integrand has a denominator that factors into distinct linear terms, partial fractions is the most direct method, and the coefficients can always be found quickly by plugging in the roots of each factor.
- The result can be checked by differentiating $\tfrac14\ln\left|\tfrac{x-1}{x+3}\right|$ with the quotient/chain rule and confirming it returns $\dfrac{1}{(x-1)(x+3)}$.
Result
The solution to the integral is: