[Largedisplaystyle int_{0}^{1}frac{sqrt[4]{xleft ( 1-x
ight )^{3}}}{left ( 1+x
ight )^{3}}mathrm{d}x~~,~~int_{0}^{1}frac{sqrt[3]{xleft ( 1-x
ight )^{2}}}{left ( 1+x
ight )^{3}}mathrm{d}x
]
(Largemathbf{Solution:})
我们来计算更一般的情况：
[Ileft ( m,n
ight )=int_{0}^{1}frac{sqrt[n]{x^{m}left ( 1-x
ight )^{n-m}}}{left ( 1+x
ight )^{3}}mathrm{d}x
]
为了化为Beta函数，我们作如下变换
[t=frac{2x}{1+x},~ ~ ~ 1-t=frac{1-x}{1+x},~ ~ ~ mathrm{d}t=frac{2mathrm{d}x}{left ( 1+x
ight )^{2}}
]
于是直接计算得：
[egin{align*}
int_{0}^{1}frac{sqrt[n]{x^{m}left ( 1-x
ight )^{n-m}}}{left ( 1+x
ight )^{3}}mathrm{d}x &=int_{0}^{1}left ( frac{x}{1+x}
ight )^{frac{m}{n}}left ( frac{1-x}{1+x}
ight )^{frac{n-m}{n}}frac{mathrm{d}x}{left ( 1+x
ight )^{2}} \
&=2^{-frac{n+m}{n}}int_{0}^{1}t^{frac{m}{n}}left ( 1-t
ight )^{frac{n-m}{n}}mathrm{d}t \
&=frac{2^{-frac{n+m}{n}}}{Gamma left ( 3
ight )}Gamma left ( frac{m+n}{n}
ight )Gamma left ( frac{2n-m}{n}
ight ) \
&=2^{-frac{2n+m}{n}}cdot frac{m}{n}cdot frac{n-m}{n}cdot Gamma left ( frac{m}{n}
ight )cdot Gamma left ( 1-frac{m}{n}
ight )\
&=2^{-frac{2n+m}{n}}cdot frac{mleft ( n-m
ight )}{n^{2}}cdot frac{pi }{sinleft ( dfrac{mpi }{n}
ight )}
end{align*}]
所以原题即为：
[Largeoxed{egin{align*}
Ileft ( 1,4
ight )&=color{Blue} {frac{3sqrt[4]{2}}{64}pi}\[15pt]
Ileft ( 1,3
ight )&=color{Blue} {frac{pi }{18}frac{sqrt[3]{4}}{sqrt{3}}}
end{align*}}]