[Largedisplaystyle int_{0}^{1}xsqrt{1+x^{3}}mathrm{d}x
]
(Largemathbf{Solution:})
易知
[int_{0}^{1}xsqrt{1+x^{3}}mathrm{d}x=frac{1}{3}int_{0}^{1}x^{-frac{1}{3}}left ( 1+x
ight )^{frac{1}{2}}mathrm{d}x
]
下面我们来看这个一般形式
[int_{0}^{u}y^{b -1}left ( u-y
ight )^{c-b-1}left ( y+frac{u}{x}
ight )^{-a}mathrm{d}t
]
首先我们引入Beta函数
[mathrm{B}left ( a,b
ight )=int_{0}^{1}t^{a-1}left ( 1-t
ight )^{b-1}mathrm{d}t
]
然后引入超几何函数 (_{2}F_{1}) 的定义
[_{2}F_{1}left ( a,b;c;x
ight )=frac{1}{mathrm{B}left ( b,c-b
ight )}int_{0}^{1}t^{b-1}left ( 1-t
ight )^{c-b-1}left ( 1-tx
ight )^{-a}mathrm{d}t
]
简单调整之后我们可以得到
[int_{0}^{1}t^{b}left ( 1-t
ight )^{c}left ( 1-tx
ight )^{a}mathrm{d}t=mathrm{B}left ( b+1,c+1
ight )\, _{2}F_{1}left (-a,b+1;b+c+2;x
ight )
]
做代换 (y=tu~,~x ightarrow -x)后，我们有
[egin{align*}
&int_{0}^{1}t^{b}left ( 1-t
ight )^{c}left ( 1-tx
ight )^{a}mathrm{d}t=int_{0}^{1}left ( frac{y}{u}
ight )^{b}left ( 1-frac{y}{u}
ight )^{c}left ( 1+frac{yx}{u}
ight )^{a}frac{1}{u}mathrm{d}y\
&=left ( frac{u}{x}
ight )^{-a}u^{-b-c-1}int_{0}^{u}y^{b}left ( u-y
ight )^{c}left ( y+frac{u}{x}
ight )^{a}mathrm{d}y
end{align*}]
然后做代换 (b+1 ightarrow b~,~c+1 ightarrow c-b~,~a ightarrow -a) 我们有
[int_{0}^{u}y^{b -1}left ( u-y
ight )^{c-b-1}left ( y+frac{u}{x}
ight )^{-a}mathrm{d}t=left ( frac{u}{x}
ight )^{a}u^{c-1}mathrm{B}left ( b,c-b
ight )\, _{2}F_{1}left ( a,b;c;-x
ight )
]
所以我们令 (u=1~,~x=1~,~b=dfrac{2}{3}~,~c=dfrac{5}{3}~,~a=-dfrac{1}{2})，可以得到
[int_{0}^{1}x^{-frac{1}{3}}left ( 1+x
ight )^{frac{1}{2}}mathrm{d}x=frac{3}{2}\, _{2}F_{1}left ( -frac{1}{2},frac{2}{3};frac{5}{3};-1
ight )
]
所以
[Largeoxed{displaystyle int_{0}^{1}xsqrt{1+x^{3}}mathrm{d}x=color{blue}{frac{1}{2}\, _{2}F_{1}left ( -frac{1}{2},frac{2}{3};frac{5}{3};-1
ight )}}
]