一道微积分题求解答 一道微积分题，求解答需要过程？
\u4e00\u9053\u5fae\u79ef\u5206\u9898\u76ee\uff0c\u6c42\u89e3\u7b54\u8bbe f(x)=\u222b[1,x] ln(1+t)/t dt \u4ee4u=1/t
=\u222b[1,1/x] uln(1+1/u) d1/u
=\u222b[1,1/x] -[ln(1+u)-lnu] / udu
=\u222b[1,1/x] -ln(1+u) / udu+ \u222b[1,1/x] lnu / udu
=-f(1/x)+\u222b[1,1/x] lnu / udu
=-f(1/x)+\u222b[1,1/x] lnu dlnu
=-f(1/x)+(lnu)^2/2 | [1,1/x]
=-f(1/x)+(ln1/x)^2/2
\u2234f(x)+f(1/x)=(ln1/x)^2/2
\u222bx^3.\u221a(4-x^2) dx
=-(1/3)\u222bx^2. d(4-x^2)^(3/2)
=-(1/3)x^2. (4-x^2)^(3/2) +(2/3)\u222bx (4-x^2)^(3/2) dx
=-(1/3)x^2. (4-x^2)^(3/2) -(2/15)\u222b d(4-x^2)^(5/2)
=-(1/3)x^2. (4-x^2)^(3/2) -(2/15)(4-x^2)^(5/2) +C
具体说，在[0，兀/2] 上， sinxcosx = sinx*√[1-(sinx)^2] ，
但在 [兀/2，兀] 上，sinxcosx = -sinx*√[1-(sinx)^2] 。
sinx=sin（π-x），cosx≠cos（π-x）
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绛旓細鐞冪殑浣撶Н锛歏=4/3蟺r³锛屽垯 dV=4蟺r²dr銆傚綋r澧炲姞杈冨皬鏃讹紝DetV=4蟺r²Detr锛孌et鎸囧鍔犻噺 =4*3.14*5²*1/16=19.62 绔嬫柟鍘樼背 绛旀鍙傝冧簬鐧惧害
