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a)Xét : $I_n=\int\limits \frac{dx}{(x^2+1)^2}=\int\limits (x^2+1)^{-1};\forall x \in Z^+$
Đặt : $\left\{ \begin{array}{l} u=(x^2+1)^{-n} \Rightarrow du=\frac{-2nxdx}{(1+x^2)^{n+1}} \\ dv=dx \Rightarrow v=x \end{array} \right. $
Lúc đó : $I_n=\frac{x}{(x^2+1)^n}+2n \int\limits \frac{x^2dx}{(x^2+1)^{n+1}}=\frac{x}{(x^n+1)^n+2n \int\limits \frac{(x^2+1)-1}{(x^2+1)^{n+1}} }dx $
$=\frac{x}{(x^2+1)^n}+2nI_n-2nI_{n+1} $
$\Rightarrow I_{n+1}=\frac{x}{2n(x^2+1)^n}+\frac{(2n-1),I_n}{2n} $
$\Rightarrow I_n =\frac{x}{2(n-1)(x^2+1)^{n-1}}+\frac{(2n-3).I_{n-1}}{2(n-1)} (1);n=2;3;4....$
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