Evaluate the integral: I = integral (x² + 3)/(x² - 1) dx
Evaluate the integral: $$ I = \int \frac{x^2 + 3}{x^2 - 1} \, dx. $$
Explanation of Method
As in the previous problem, the numerator and denominator have equal degree, so the fraction is improper and should be split into a constant plus a proper fraction first. The proper fraction $\frac{1}{x^2-1}$ then factors as $\frac{1}{(x-1)(x+1)}$ and can be decomposed into partial fractions, each of which integrates to a logarithm.
- Rewrite the improper fraction as a constant plus a proper fraction.
- Factor the denominator of the remaining fraction as a difference of squares.
- Decompose into partial fractions.
- Integrate each term and combine, adding the constant of integration.
Step 1 · Decompose the fraction
The integrand can be rewritten by splitting $\dfrac{x^2+3}{x^2-1}$ as follows:
$$ \frac{x^2+3}{x^2-1} = 1 + \frac{4}{x^2-1}. $$
Thus, the integral becomes:
$$ I = \int \frac{x^2+3}{x^2-1}\,dx = \int 1\,dx + \int \frac{4}{x^2-1}\,dx. $$
Step 2 · Integrate each term
First term: $\displaystyle \int 1\,dx$:
$$ \int 1\,dx = x. $$
Second term: $\displaystyle \int \frac{4}{x^2-1}\,dx$. The term $\dfrac{1}{x^2-1}$ can be decomposed using partial fractions:
$$ \frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right). $$
Therefore:
$$ \frac{4}{x^2-1} = 2\left(\frac{1}{x-1} - \frac{1}{x+1}\right). $$
Substituting this back, we get:
$$ \int \frac{4}{x^2-1}\,dx = \int 2\left(\frac{1}{x-1} - \frac{1}{x+1}\right) dx. $$
Now, integrate each term:
$$ \int \frac{4}{x^2-1}\,dx = 2\ln|x-1| - 2\ln|x+1| = 2\ln\left|\frac{x-1}{x+1}\right|. $$
Step 3 · Combine the results
Combining the results of the two terms, we have:
$$ I = x + 2\ln\left|\frac{x-1}{x+1}\right| + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Type | Antiderivative |
|---|---|---|
| $1$ | Constant term | $x$ |
| $4/(x^2-1)$ | Partial fractions | $2\ln\left|\dfrac{x-1}{x+1}\right|$ |
Additional Notes
- The domain of $I$ requires $x \neq \pm 1$, since the denominator $x^2-1$ vanishes there.
- Partial fraction decomposition of $\dfrac{1}{(x-1)(x+1)}$ into $\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)$ is found by matching coefficients or the cover-up method.
- The result can be checked by differentiating $x + 2\ln\left|\dfrac{x-1}{x+1}\right|$ and confirming it returns the original integrand.
Result
The solution to the integral is: