Evaluate the integral: I = integral (x^4+1)/(x^6+1) dx

Evaluate the integral: $$ I = \int \frac{x^4+1}{x^6+1}\,dx. $$

Step 1 · Factor and split

$$ x^6+1=(x^2+1)(x^4-x^2+1). $$

The integrand splits as:

$$ \frac{x^4+1}{x^6+1} = \frac{2/3}{x^2+1} + \frac13\cdot\frac{x^2+1}{x^4-x^2+1}. $$

This can be verified directly: putting the right side over the common denominator $x^6+1$ gives numerator $2(x^4-x^2+1)+(x^2+1)^2=2x^4-2x^2+2+x^4+2x^2+1=3x^4+3=3(x^4+1)$, so dividing by $3$ recovers the left side exactly.

Step 2 · Integrate the first piece

$$ \frac23\int\frac{dx}{x^2+1} = \frac23\arctan x. $$

Step 3 · Integrate the second piece via $t=x-1/x$

Divide numerator and denominator of $\dfrac{x^2+1}{x^4-x^2+1}$ by $x^2$:

$$ \frac{x^2+1}{x^4-x^2+1} = \frac{1+1/x^2}{x^2-1+1/x^2}. $$

Let $t=x-\dfrac1x$, so $dt=\left(1+\dfrac1{x^2}\right)dx$ and $t^2=x^2-2+\dfrac1{x^2}$, giving $x^2-1+\dfrac1{x^2}=t^2+1$. Then:

$$ \frac13\int\frac{x^2+1}{x^4-x^2+1}\,dx = \frac13\int\frac{dt}{t^2+1} = \frac13\arctan t = \frac13\arctan\left(\frac{x^2-1}{x}\right). $$

Step 4 · Combine

$$ I = \frac23\arctan x + \frac13\arctan\left(\frac{x^2-1}{x}\right) + C. $$

Summary Table

Additional Notes

• Correction: the source's derivation was self-contradictory — it introduced inconsistent intermediate partial-fraction claims that did not reproduce the original integrand under a common-denominator check. The derivation above instead uses the verified split $f(x)=\tfrac23/(x^2+1)+\tfrac13(x^2+1)/(x^4-x^2+1)$, confirmed by direct algebraic recombination, and the standard $t=x-1/x$ substitution (the same technique used for $\int(x^2+1)/(x^4+kx^2+1)\,dx$-type integrals elsewhere in this collection).
• The domain is all real $x$, since $x^6+1>0$ everywhere.
• The result can be checked by differentiating $\tfrac23\arctan x+\tfrac13\arctan\left(\tfrac{x^2-1}x\right)$: using $\tfrac{d}{dx}\arctan\left(\tfrac{x^2-1}x\right)=\tfrac{x^2+1}{x^4-x^2+1}$, the sum is $\tfrac{2}{3(x^2+1)}+\tfrac{x^2+1}{3(x^4-x^2+1)}$, which combines over $x^6+1$ to $\tfrac{2(x^4-x^2+1)+(x^2+1)^2}{3(x^6+1)}=\tfrac{3(x^4+1)}{3(x^6+1)}=\tfrac{x^4+1}{x^6+1}$, matching the integrand exactly.