请问这道高等数学不定积分题怎么写？ 高等数学，请问这道题怎么求不定积分?
\u9ad8\u7b49\u6570\u5b66\uff0c\u8bf7\u6559\u8fd9\u9053\u4e0d\u5b9a\u79ef\u5206\u600e\u4e48\u7b97?\u539f\u5f0f=\u222bx/(x²+1)dx+\u222b1/(x²+1)dx
=1/2\u00b7\u222b1/(x²+1)\u00b7d(x²+1)+arctanx
=1/2\u00b7ln(x²+1)+arctanx+C
\u3010\u9644\u6ce8\u3011
\u222b1/(x²+1)dx=arctanx+C
\u222bx/(x²+1)dx=1/2\u00b7ln(x²+1)+C
\u539f\u5f0f=-\u222b1/\u221a(1+x²)*d(1/x)
\u4ee41/x=t,\u5219x=1/t
\u539f\u5f0f=-\u222bdt/\u221a(1+1/t²)
=-\u222btdt/\u221a(t²+1)
=-1/2*\u222b(t²+1)^(-1/2)*d(t²+1)
=-(t²+1)^(1/2)+C
=-\u221a(1+1/x²)+C
=∫(0->π) √(sin^2 x/2 +cos^2 x/2 -2sinx/2 * cosx/2) dx
=∫(0->π) √(sinx/2 -cosx/2)^2 dx
∵0<=x<=π
∴0<=x/2<=π/2
当0<=x/2<=π/4时
cosx/2>sinx/2
当π/4<=x/2<=π/2时
cosx/2<sinx/2
∴ 原式=∫(0->π/2) (cosx/2 -sinx/2) dx +∫(π/2->π) (sinx/2 -cosx/2) dx
=∫(0->π/2) cosx/2dx-∫(0->π/2) sinx/2 dx +∫(π/2->π) sinx/2 dx -∫(π/2->π) cosx/2 dx
=2 sinx/2|(0->π/2) +2cosx/2|(0->π/2) -2cosx/2|(π/2->π) -2sinx/2|(π/2->π)
=2(√2/2 -0) +2(√2/2 -1) -2(0-√2/2) -2(1-√2/2)
=√2 +√2-2 +√2 -2+√2
=4√2-4
你是电气几班
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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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绛旓細涓銆佲垰(1 + x²) = 鈭(1 + tan²y) = 鈭歴ec²y = secy 璁緓 = tany锛宒x = sec²y dy 鈭 x⁵/鈭(1 + x²) dx = 鈭 tan⁵ysec²y/secy dy = 鈭 tan⁵ysecy dy = 鈭 (sec²y - 1)² dsecy = 鈭 (...
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