求幂级数\[\sum_{n=0}^{+\infty}\dfrac{n^3-5n+1}{(2n+1)(n+2)}x^{2n+2}\]的和函数 $S(x)$．
答案 $\dfrac{11x^4-13x^2+4}{4(x^2-1)^2}+\dfrac 94x\mathop{\rm arctan}x+\dfrac{\ln(1-x^2)}{x^2}$．
解析 根据题意，有\[\begin{split} \sum_{n=0}^{+\infty}\dfrac{n^3-5n+1}{(2n+1)(n+2)}x^{2n+2}&=\sum_{n=0}^{+\infty}\left(-\dfrac 54+\dfrac n2-\dfrac{1}{n+2}+\dfrac{9}{4(2n+1)}\right)x^{2n+2}\\ &=-\dfrac{5x^2}{4(1-x^2)}+\dfrac{x^4}{2(x^2-1)^2}-\left(-1-\dfrac{\ln(1-x^2)}{x^2}\right)+\dfrac 94x\mathop{\rm arctanh} x\\ &=\dfrac{11x^4-13x^2+4}{4(x^2-1)^2}+\dfrac 94x\mathop{\rm arctan}x+\dfrac{\ln(1-x^2)}{x^2}.\end{split}\]