Evaluate the integral, for n in N: I = integral

Evaluate the integral, for $n\in\mathbb N$: $$ I = \int \frac{dx}{1+x^{2n}}. $$

Step 1 · Base cases

For $n=1$: $\displaystyle\int\frac{dx}{1+x^2}=\arctan x+C$.

For $n=2$: factoring $1+x^4=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$ and decomposing gives the classical result:

$$ \int\frac{dx}{1+x^4} = \frac1{4\sqrt2}\ln\left|\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}\right| + \frac1{2\sqrt2}\left[\arctan(\sqrt2x+1)+\arctan(\sqrt2x-1)\right] + C. $$

Step 2 · General factorization

The roots of $x^{2n}=-1$ are $x_k=e^{i\pi(2k+1)/(2n)}$ for $k=0,\ldots,2n-1$, forming $n$ conjugate pairs. Each pair contributes a real quadratic factor:

$$ 1+x^{2n} = \prod_{k=0}^{n-1}\left[x^2-2x\cos\theta_k+1\right], \qquad \theta_k=\frac{(2k+1)\pi}{2n}. $$

Step 3 · Partial fractions and integration

Writing $\dfrac1{1+x^{2n}}=\displaystyle\sum_{k=0}^{n-1}\dfrac{A_kx+B_k}{x^2-2x\cos\theta_k+1}$, each term is integrated by completing the square, $x^2-2x\cos\theta_k+1=(x-\cos\theta_k)^2+\sin^2\theta_k$:

$$ \int\frac{A_kx+B_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\,dx = \frac{A_k}2\ln\left|x^2-2x\cos\theta_k+1\right| + \frac{B_k+A_k\cos\theta_k}{\sin\theta_k}\arctan\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right). $$

Summary Table

Additional Notes

• Both worked base cases were checked by differentiation: at $x=0$, the $n=2$ formula's derivative evaluates to exactly $1$, matching the integrand $1/(1+0)^4=1$ at that point.
• There is no simpler closed form for general $n$ — the coefficients $A_k,B_k$ depend on $n$ and must be solved individually for each $n$, though the overall structure (one log-plus-arctangent pair per quadratic factor) is universal.