Evaluate the integral: I = integral (x²+1)/(x^4+1) dx
Evaluate the integral: $$ I = \int \frac{x^2+1}{x^4+1} \, dx. $$
Explanation of Method
Dividing numerator and denominator by $x^2$ reveals a hidden structure: the denominator becomes $x^2+\tfrac1{x^2}$, which equals $\left(x-\tfrac1x\right)^2+2$, while the numerator becomes exactly the differential of $t=x-\tfrac1x$. This classic substitution turns the whole integral into a standard arctangent form.
- Divide numerator and denominator by $x^2$.
- Substitute $t=x-\tfrac1x$ and recognize $x^2+\tfrac1{x^2}=t^2+2$.
- Apply the standard arctangent formula.
- Substitute back to express the answer in terms of $x$.
Step 1 · Substitute $t = x - 1/x$
We use the substitution $t = x - \dfrac{1}{x}$, so that:
$$ dt = \left(1+\frac{1}{x^2}\right) dx. $$
Dividing the numerator and denominator of the original integrand by $x^2$ gives:
$$ \frac{x^2+1}{x^4+1} = \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}, $$
and since $x^2+\dfrac{1}{x^2} = \left(x-\dfrac1x\right)^2+2 = t^2+2$, the numerator matches $dt$ exactly.
Step 2 · Rewrite the integral in terms of $t$
Rewriting the integral in terms of $t$, we get:
$$ I = \int \frac{dt}{t^2+2}. $$
Step 3 · Apply the standard result
This is a standard integral:
$$ \int \frac{dt}{t^2+a^2} = \frac{1}{a}\arctan\frac{t}{a}. $$
Using $a = \sqrt2$, we obtain:
$$ I = \frac{1}{\sqrt{2}}\arctan\frac{t}{\sqrt2} + C. $$
Step 4 · Substitute back
Substituting back $t = x - \dfrac1x = \dfrac{x^2-1}{x}$, we obtain:
$$ I = \frac{1}{\sqrt{2}}\arctan\frac{x^2-1}{x\sqrt2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $t = x - 1/x$ |
| $x^2+1/x^2$ in terms of $t$ | $t^2+2$ |
| Antiderivative | $\tfrac{1}{\sqrt2}\arctan(t/\sqrt2)$ |
Additional Notes
- The domain of $I$ requires $x \neq 0$, since the substitution $t=x-1/x$ involves dividing by $x$.
- The substitution $t=x-\tfrac1x$ (or $t=x+\tfrac1x$ for a related family of integrals) is a standard technique whenever an integrand is symmetric under $x \to 1/x$ up to a sign — here, dividing through by $x^2$ exposes exactly that symmetry.
- The result can be checked by differentiating $\tfrac{1}{\sqrt2}\arctan\left(\tfrac{x^2-1}{x\sqrt2}\right)$ and confirming (after simplification) that it returns $\dfrac{x^2+1}{x^4+1}$.
Result
The solution to the integral is: