= Solution
Use the <Maclaurin series> of the <logarithm> and <hyperbolic cosine>:
$$
\ln(1+x^3)=x^3-\frac12x^6+O(x^9),\qquad
\frac1{\cosh x}=1-\frac12x^2+\frac5{24}x^4+O(x^6).
$$
To retain terms through degree six, only the constant and quadratic terms of the reciprocal factor can multiply the leading cubic term. Thus
$$
\boxed{\frac{\ln(1+x^3)}{\cosh x}=x^3-\frac12x^5-\frac12x^6+O(x^7).}
$$
The numerator in the original PDF is $\ln(1+x^3)$; the cubic power is missing from the local TeX transcription.
Back to article page
