Question

Evaluate: \[ \int \frac{1}{x(x^6 + 1)} \, dx \]

Hint:

For integrals involving a term like \( x(x^n+k) \), the substitution \( u=x^n \) is often effective after multiplying the numerator and denominator by \( x^{n-1} \).

Answer 1

Step 1: Multiply the numerator and denominator by \( x^5 \) to prepare for a substitution. \[ \int \frac{x^5}{x^6(x^6 + 1)} dx \] Step 2: Let \( u = x^6 \). Then \( du = 6x^5 dx \), which implies \( x^5 dx = \frac{du}{6} \).
Step 3: Substitute \(u\) and \(du\) into the integral. \[ \int \frac{1}{u(u+1)} \frac{du}{6} = \frac{1}{6} \int \frac{1}{u(u+1)} du \] Step 4: Use partial fraction decomposition for the integrand. \[ \frac{1}{u(u+1)} = \frac{A}{u} + \frac{B}{u+1} \implies 1 = A(u+1) + Bu \] If \( u=0 \), \( 1 = A \). If \( u=-1 \), \( 1 = -B \implies B=-1 \).
So, \( \frac{1}{u(u+1)} = \frac{1}{u} - \frac{1}{u+1} \).
Step 5: Integrate the resulting expression. \[ \frac{1}{6} \int \left(\frac{1}{u} - \frac{1}{u+1}\right) du = \frac{1}{6} (\ln|u| - \ln|u+1|) + C \] Step 6: Simplify using logarithm properties and substitute back \( u=x^6 \). \[ \frac{1}{6} \ln\left|\frac{u}{u+1}\right| + C = \frac{1}{6} \ln\left|\frac{x^6}{x^6+1}\right| + C \]