Evaluate the integral: I = integral ((2-x)²)/(2-x²) dx
Evaluate the integral: $$ I = \int \frac{(2-x)^2}{2-x^2}\,dx. $$
Explanation of Method
The numerator and denominator have the same degree, so polynomial long division first isolates a constant term, leaving a proper rational function that splits by partial fractions into two simple logarithmic pieces — one from a substitution and one from the standard $\int \frac{dx}{x^2-a^2}$ formula.
- Expand the numerator.
- Divide the resulting polynomial by the denominator.
- Split the remainder into a substitution term and a standard $1/(x^2-a^2)$ term.
- Integrate each piece and combine.
Step 1 · Expand the numerator
Expanding $(2-x)^2$:
$$ (2-x)^2 = 4 - 4x + x^2. $$
Thus, the integral becomes:
$$ I = \int \frac{x^2-4x+4}{2-x^2}\,dx. $$
Step 2 · Polynomial division
Both numerator and denominator are degree 2, so we divide. Writing the denominator as $2-x^2=-(x^2-2)$:
$$ \frac{x^2-4x+4}{x^2-2} = 1 + \frac{-4x+6}{x^2-2}, $$
since $x^2-4x+4 = 1\cdot(x^2-2) + (-4x+6)$. Therefore:
$$ \frac{x^2-4x+4}{2-x^2} = -\left(1+\frac{-4x+6}{x^2-2}\right) = -1 + \frac{4x-6}{x^2-2}. $$
So the integral splits as:
$$ I = \int \left(-1 + \frac{4x}{x^2-2} - \frac{6}{x^2-2}\right) dx. $$
Step 3 · Integrate each piece
First piece: $\displaystyle\int -1\,dx = -x$.
Second piece: for $\displaystyle\int \frac{4x}{x^2-2}\,dx$, let $w=x^2-2$, so $dw=2x\,dx$:
$$ \int \frac{4x}{x^2-2}\,dx = 2\int \frac{dw}{w} = 2\ln|x^2-2|. $$
Third piece: using the standard form $\displaystyle\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|$ with $a=\sqrt2$:
$$ -6\int \frac{dx}{x^2-2} = -6\cdot\frac{1}{2\sqrt2}\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right| = -\frac{3\sqrt2}{2}\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right|. $$
Step 4 · Combine the results
Adding all three pieces together:
$$ I = -x + 2\ln|x^2-2| - \frac{3\sqrt2}{2}\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right| + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $-1$ | $-x$ |
| $4x/(x^2-2)$ | $2\ln|x^2-2|$ |
| $-6/(x^2-2)$ | $-\tfrac{3\sqrt2}{2}\ln\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|$ |
Additional Notes
- The domain excludes $x=\pm\sqrt2$, where the original denominator $2-x^2$ vanishes.
- A common pitfall here is to factor $2-x^2$ as $-(2-x)(2+x)$, but $(2-x)(2+x)=4-x^2$, not $2-x^2$ — the correct factorization is $2-x^2=-(x-\sqrt2)(x+\sqrt2)$, which is what drives the $\ln\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|$ term above.
- The result can be checked by differentiating each piece: $\tfrac{d}{dx}\left[2\ln|x^2-2|\right]=\tfrac{4x}{x^2-2}$ and $\tfrac{d}{dx}\left[-\tfrac{3\sqrt2}{2}\ln\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|\right]=\tfrac{-6}{x^2-2}$, which together with $-1$ reconstruct the original integrand.
Result
The solution to the integral is: