这题高等数学的不定积分怎么做
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仅供参考
分部积分：
∫ln(1+x^2)dx
=xln(1+x^2)-∫2x^2/(1+x^2)dx
=xln(1+x^2)-[∫2dx-∫2/(1+x^2)dx]
=xln(1+x^2)-2x+2arctanx+C
奇函数，定积分偶倍奇零，=0
绛旓細=鈭(x-1)/(x^2-x+1)dx-鈭1/(x^2-x+1)dx =1/2ln(x^2-x+1)-鈭1/(x-1/2)^2+3/4)dx =1/2ln(x^2-x+1)-4/3*鈭1/(2x/鈭3-1/鈭3)^2+1)dx =1/2ln(x^2-x+1)-2/鈭3*arctan(2x/鈭3-1/鈭3)+C ...
绛旓細鈭(0->蟺) 鈭(1-sinx) dx =鈭(0->蟺) 鈭(sin^2 x/2 +cos^2 x/2 -2sinx/2 * cosx/2) dx =鈭(0->蟺) 鈭(sinx/2 -cosx/2)^2 dx 鈭0<=x<=蟺 鈭0<=x/2<=蟺/2 褰0<=x/2<=蟺/4鏃 cosx/2>sinx/2 褰撓/4<=x/2<=蟺/2鏃 cosx/2<sinx/2 鈭 鍘熷紡=鈭(0...
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绛旓細鍥炵瓟锛氭牴鎹棰樼洰鏈夆埆f(x)dx= ln x/x 涓よ竟瀵箈姹傚鏈 f(x)=(1-ln x)/x² 鍘熷紡=鈭玿 d f(x) =x f(x) - 鈭玣(x) dx =(1-ln x)/x -ln x/x =(1-2ln x)/x
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绛旓細濡傚浘鎵绀猴細
绛旓細鍘熷紡=鈭玿/(x²+1)dx+鈭1/(x²+1)dx =1/2路鈭1/(x²+1)路d(x²+1)+arctanx =1/2路ln(x²+1)+arctanx+C 銆愰檮娉ㄣ戔埆1/(x²+1)dx=arctanx+C 鈭玿/(x²+1)dx=1/2路ln(x²+1)+C ...
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绛旓細濡傚浘
