高等数学，不定积分问题，求解题思路与步骤 高等数学不定积分问题高数换元积分法
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\u62c6\u5f00\u6765\u5206\u522b\u6c42\u5c31\u53ef\u4ee5\u4e86
\u3000\u3000\u89e3\uff1a\u2460\u7c7b\u4f3c\u4e8e26\u300127\u300128\u9898\u8fd9\u79cd\u7c7b\u578b\u7684\u9898\u76ee\uff0c\u4e00\u822c\u53ef\u4e0d\u8fdb\u884c\u6362\u5143\uff0c\u800c\u662f\u901a\u8fc7\u4e09\u89d2\u51fd\u6570\u4e2d\u7684\u201c\u79ef\u5316\u548c\u5dee\u201d\u516c\u5f0f\uff0c\u8f6c\u5316\u6210\u6b63\u5f26\u6216\u8005\u4f59\u5f26\u51fd\u6570\uff0c\u5373\u53ef\u6c42\u89e3\u3002
\u3000\u3000\u2461\u7c7b\u4f3c\u4e8e31\u300132\u9898\u7684\u9898\u76ee\uff0c\u8f83\u7b80\u5355\u7684\uff0c\u53ef\u4ee5\u7528\u201c\u51d1\u201d\u79ef\u5206\u65b9\u6cd5\u89e3\u51b3\uff0c\u5982\u56fe\u7247\u4e2d\u7684\u89e3\u6cd5\u5373\u662f\u3002\u4ea6\u53ef\u7528\u201c\u53bb\u6839\u53f7\u221a\u3001\u53bb\u5206\u6bcd\u201d\u7684\u6362\u5143\u6cd5\u6c42\u89e3\u3002\u598231\u9898\uff0c\u53ef\u8bbex=(3/2)sint\uff0c\u53bb\u6839\u53f7\u6c42\u89e3\uff1b32\u9898\uff0c\u53ef\u8bbex=3tan\u03b8\uff0c\u53bb\u5206\u6bcd\u800c\u6c42\u89e3\u3002
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=1/2*∫(2x+2)dx/(x²+2x+3)-1/2*∫4dx/(x²+2x+3)
=1/2*∫d(x²+2x+3)/(x²+2x+3)-2∫d(x+1)/[(x+1)²+2]
=1/2*ln|x²+2x+3|-√2*arctan[(x+1)/√2]+C
[ln(x^2+2x+3) ]'= (2 x + 2) / (x^2 + 2 x + 3)
[arctan(x)]' = 1/ ( 1 + x^2)
[arctan( (x+1) / a )]' = 1/ [ a *( 1 + ((x+1)/a)^2))]
积分 =0.5* ln(x^2+2x+3) - a * arctan( (x+1) / a ) +C
a = Sqrt(2)
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绛旓細1銆佸師寮=鈭4/(2sinxcosx*2cos^2x)dx =鈭4/[sin2x*(1+cos2x)]dx 浠=tanx锛宻in2x=2u/(1+u^2)锛宑os2x=(1-u^2)/(1+u^2)锛宒x=du/(1+u^2)鍘熷紡=鈭4/[2u/(1+u^2)*2/(1+u^2)]*du/(1+u^2)=鈭(1+u^2)/udu =鈭(1/u+u)du =ln|u|+u^2/2+C =ln|tanx|...
绛旓細2. f(x) = {ln[x+鈭(1+x^2)]}' = [1+x/鈭(1+x^2)]/[x+鈭(1+x^2)] = 1/鈭(1+x^2)鍒 鈭玿f'(x)dx = 鈭玿df(x) = xf(x) - 鈭玣(x)dx = x/鈭(1+x^2) - ln[x+鈭(1+x^2)] + C
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绛旓細鍘熷紡=-鈭玞os2xde^(-x)=-(cos2x)e^(-x)+鈭玡^(-x)d(cos2x)=-(cos2x)e^(-x)-2鈭玡^(-x) sin2x dx =-(cos2x)e^(-x)+2鈭 sin2x de^(-x)=-(cos2x)e^(-x)+2 (sin2x) e^(-x) - 2 鈭 e^(-x) d(sin2x)=-(cos2x)e^(-x)+2 (sin2x) e^(-x) - 4 鈭...
绛旓細鍘熷紡=鈭玠x/x鈭歔(x+3/2)^2-25/4]浠+3/2=(5/2)*sect锛宒x=(5/2)*secttantdt 鍘熷紡=鈭(5/2)*secttantdt/{[(5/2)*sect-(3/2)]*(5/2)*tant} =鈭2sectdt/(5sect-3)=鈭2dt/(5-3cost)浠=tan(t/2)锛屽垯cost=(1-u^2)/(1+u^2)锛宒t=2du/(1+u^2)鍘熷紡=鈭玔4...
