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

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

Explanation of Method

Unlike Problem 276, the roots of $x^{2k}=1$ include two real values, $x=\pm1$, in addition to complex-conjugate pairs. The partial fraction decomposition therefore has two linear-factor log terms plus one log/arctangent pair for each remaining quadratic factor.

Verify the pattern on the elementary cases $k=1$ and $k=2$.
Factor $1-x^{2k}$ over the reals.
Decompose into partial fractions.
Integrate each term.

Step 1 · Base cases

For $k=1$:

$$ \int\frac{dx}{1-x^2} = \frac12\ln\left|\frac{1+x}{1-x}\right| + C. $$

For $k=2$: factoring $1-x^4=(1-x)(1+x)(1+x^2)$ and solving $\tfrac1{1-x^4}=\tfrac{1/4}{1-x}+\tfrac{1/4}{1+x}+\tfrac{1/2}{1+x^2}$ gives:

$$ \int\frac{dx}{1-x^4} = \frac14\ln\left|\frac{1+x}{1-x}\right| + \frac12\arctan x + C. $$

Step 2 · General factorization

Since $1-x^{2k}=(1-x^k)(1+x^k)$ and its $2k$ roots are the $2k$-th roots of unity, the real factorization always contains the two linear factors $(1-x)$ and $(1+x)$, together with $(k-1)$ real quadratic factors from the remaining conjugate root pairs.

Step 3 · Partial fractions and integration

The linear factors contribute logarithms; each quadratic factor $x^2-2x\cos\theta_j+1$ (for the appropriate angles $\theta_j=\pi j/k$) contributes a log/arctangent pair exactly as in Problem 276.

Summary Table

$k$	Result
$1$	$\tfrac12\ln\left\|\tfrac{1+x}{1-x}\right\|=\operatorname{artanh}(x)$
$2$	$\tfrac14\ln\left\|\tfrac{1+x}{1-x}\right\|+\tfrac12\arctan x$
general	Two logs from $x=\pm1$, plus one log/arctan pair per quadratic factor

Additional Notes

Both base cases were checked by differentiation at $x=0.5$: the $k=2$ formula's derivative evaluates to $1.0\overline{6}$, matching $1/(1-0.5^4)=1/0.9375\approx1.0667$.
As with Problem 276, the exact coefficients depend on $k$ and must be solved case by case; the overall structure (two simple logarithms plus $k-1$ log/arctangent pairs) is universal.

Result

The general solution is:

$$ I = \frac1{2k}\sum_{\text{real roots }r=\pm1} A_r\ln|x-r| + \sum_{\text{quadratic factors}}\left[B_j\ln\left|x^2-2x\cos\theta_j+1\right| + C_j\arctan\left(\frac{x-\cos\theta_j}{\sin\theta_j}\right)\right] + C, $$

where $\theta_j=\pi j/k$ for the appropriate range of $j$, and the constants $A_r,B_j,C_j$ come from the partial-fraction decomposition for the specific $k$.