Step 1: Verifying the division remainder differently, via synthetic-style matching:
Write \(x^4=(x-1)(x^2+1)Q(x)+R(x)\) with \(\deg R<3\). Assume \(Q(x)=x+1\) (matching leading terms) and expand \((x-1)(x^2+1)(x+1)=(x^2-1)(x+1)\cdot\dots\) — more directly, \((x-1)(x^2+1)=x^3-x^2+x-1\), and \(x^4-(x+1)(x^3-x^2+x-1)=x^4-(x^4-x^3+x^2-x+x^3-x^2+x-1)=x^4-x^4+1=1\), confirming the remainder is exactly \(1\).
Step 2: Redoing the partial fraction split via the cover-up method:
For the \(A/(x-1)\) term, 'cover up' \((x-1)\) in \(1/[(x-1)(x^2+1)]\) and evaluate the rest at \(x=1\): \(A=1/(1^2+1)=1/2\), matching the earlier result directly without solving simultaneous equations.
Step 3: Integrating each resulting piece independently:
\(\int(x+1)dx=x^2/2+x\); \(\int\frac{1/2}{x-1}dx=\frac12\ln|x-1|\); and splitting \(-\frac{x+1}{2(x^2+1)}=-\frac{x}{2(x^2+1)}-\frac{1}{2(x^2+1)}\) integrates to \(-\frac14\ln(x^2+1)-\frac12\tan^{-1}x\).
Final Answer:
\[ \boxed{\dfrac{x^{2}}{2}+x+\dfrac12\ln|x-1|-\dfrac14\ln(x^{2}+1)-\dfrac12\tan^{-1}x+C} \]