(1)
∫{\(x^{2}\)/(x+1)}dx
=∫{(\(x^{2}\)-1)/(x+1)}dx+∫{1/(x+1)}dx
=∫(x-1)dx+∫{1/(x+1)}dx
=\(x^{2}\)/2-x+log(x+1)+C
Cは積分定数

(2)
\(I_{n}\)=∫_0^∞(\(x^{n}\))*{e^(-x)}dxと置く
\(I_{n}\)=n*I_(n-1)を示す。
\(I_{n}\)
=∫_0^∞(\(x^{n}\))*{e^(-x)}dx
=∫_0^∞(\(x^{n}\))*{-e^(-x)}'dx
=[-(\(x^{n}\))e^(-x)]_0^∞+∫_0^∞{(\(x^{n}\))}'*{e^(-x)}
=n∫_0^∞{x^(n-1)}*{e^(-x)}=n*I_(n-1)

したがって、
\(I_{n}\)
=n*I_(n-1)=n*(n-1)*I_(n-2)
=…
=n!*\(I_{1}\)

\(I_{1}\)
=∫_0^∞(x)*{e^(-x)}dx
=∫_0^∞(x)*{-e^(-x)}'dx
=[-(x)e^(-x)]_0^∞+∫_0^∞{(x)}'*{e^(-x)}
=∫_0^∞{e^(-x)}=[-e^(-x)]_0^∞=1だから

\(I_{n}\)
=∫_0^∞(\(x^{n}\))*{e^(-x)}dx
=n!
となる。