Evaluate the integral: I = integral

Evaluate the integral: $$ I = \int \frac{x^5+x^4-2}{x^4-4x^2+4}\,dx. $$

Step 1 · Polynomial long division

Dividing $x^5+x^4-2$ by $x^4-4x^2+4$:

$$ \frac{x^5+x^4-2}{x^4-4x^2+4} = x+1 + \frac{4x^3+4x^2-4x-6}{x^4-4x^2+4}. $$

Step 2 · Factor the denominator and decompose

$$ x^4-4x^2+4 = (x^2-2)^2, \qquad \frac{4x^3+4x^2-4x-6}{(x^2-2)^2} = \frac{Ax+B}{x^2-2} + \frac{Cx+D}{(x^2-2)^2}. $$

Multiplying through: $4x^3+4x^2-4x-6=(Ax+B)(x^2-2)+Cx+D$. Matching coefficients: $A=4$, $B=4$, and then $-2A+C=-4\Rightarrow C=4$, $-2B+D=-6\Rightarrow D=2$.

Step 3 · Integrate the pieces

$$ \int(x+1)\,dx = \frac{x^2}2+x. $$

$$ \int \frac{4x+4}{x^2-2}\,dx = 2\ln\left|x^2-2\right| + 4\int\frac{dx}{x^2-2} = 2\ln\left|x^2-2\right| + \sqrt2\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right|. $$

For $\displaystyle\int\frac{4x+2}{(x^2-2)^2}\,dx$, split into $4\int\frac{x}{(x^2-2)^2}dx+2\int\frac{dx}{(x^2-2)^2}$. The first piece is $-\dfrac{2}{x^2-2}$. The second uses the reduction formula $\displaystyle\int\frac{dx}{(x^2-a^2)^2}=-\frac{x}{2a^2(x^2-a^2)}-\frac{1}{4a^3}\ln\left|\frac{x-a}{x+a}\right|$ with $a^2=2$:

$$ 2\int\frac{dx}{(x^2-2)^2} = -\frac{x}{2(x^2-2)} - \frac{\sqrt2}{8}\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right|. $$

Step 4 · Combine everything

$$ I = \frac{x^2}2+x + 2\ln\left|x^2-2\right| + \sqrt2\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right| - \frac2{x^2-2} - \frac{x}{2(x^2-2)} - \frac{\sqrt2}8\ln\left|\frac{x-\sqrt2}{x+\sqrt2}\right| + C. $$

Combining the two logarithms of $\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|$: $\sqrt2-\tfrac{\sqrt2}8=\tfrac{7\sqrt2}8$. Combining the two fractional terms over $x^2-2$: $-\tfrac{2}{x^2-2}-\tfrac{x}{2(x^2-2)}=-\tfrac{x+4}{2(x^2-2)}$.

Summary Table

Additional Notes

• Correction: the source's computation of $\int dx/(x^2-2)^2$ applied the reduction formula with an incorrect overall sign, effectively adding the $\sqrt2/8\,\ln|\cdots|$ term instead of subtracting it. This shifted the coefficient of $\ln\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|$ from the correct $7\sqrt2/8$ to an incorrect value. The version above rebuilds this piece from the standard reduction formula and verifies it by direct differentiation.
• The domain excludes $x=\pm\sqrt2$, where the denominator vanishes.
• The result can be checked by differentiating the boxed expression: the $x^2/2+x$ term gives $x+1$; the logarithmic terms give $\tfrac{4x+4}{x^2-2}$ after combining, matching Step 2; and $-\tfrac{d}{dx}\left[\tfrac{x+4}{2(x^2-2)}\right]-\tfrac{\sqrt2}8\cdot\tfrac{d}{dx}\ln\left|\tfrac{x-\sqrt2}{x+\sqrt2}\right|$ reproduces $\tfrac{4x+2}{(x^2-2)^2}$, so the sum matches the original integrand after long division.