求不定积分
\u600e\u4e48\u8ba1\u7b97\u4e0d\u5b9a\u79ef\u5206\u8fd9\u4e2a\u662f\u5178\u578b\u7684\u6362\u5143\u6cd5\u79ef\u5206
\u867d\u7136\u65b9\u6cd5\u8bf4\u8d77\u6765\u5f88\u5bb9\u6613\uff0c\u4f46\u662f\u80fd\u4e0d\u80fd\u505a\u51fa\u6765\u8fd8\u662f\u8981\u770b\u4f60\u5bf9\u5bfc\u6570\u5f62\u5f0f\u7684\u719f\u7ec3\u7a0b\u5ea6
\u6bd4\u5982\u8fd9\u4e00\u9898\uff0c\u5982\u679c\u4f60\u80fd\u770b\u5230e^x\u5c31\u7acb\u5373\u60f3\u5230\u5c06e^x\u653e\u5230d\u7684\u540e\u9762\uff0c\u56e0\u4e3ade^x=e^xdx
\u518d\u6bd4\u5982\uff0c\u4f60\u770b\u5230\u4e86\u222bsinxcosxdx\uff0c\u4f60\u5c31\u5e94\u8be5\u7acb\u5373\u60f3\u5230(sinx)'=cosx\uff0c\u7136\u540e\u5c06cosx\u6362\u6210sinx\u653e\u5230d\u7684\u540e\u9762\uff1a
\u222bsinxdsinx=(sinx)^2/2
\u4e00\u4e2a\u7406\u6027\u7684\u65b9\u6cd5\u5c31\u662f\uff0c\u5148\u641e\u6e05\u695a\u6362\u5143\u6cd5\u7684\u57fa\u672c\u516c\u5f0f\u548c\u65b9\u6cd5
\u7136\u540e\u591a\u505a\u6c42\u5bfc\u548c\u79ef\u5206\u7684\u7ec3\u4e60\u9898\uff0c\u591a\u505a\u603b\u7ed3\u3002\u505a\u5230\u540e\u6765\u4f60\u5c31\u4f1a\u53d1\u73b0\u5f88\u7b80\u5355\u4e86
\u8865\u5145\u4e00\u4e0b\uff1a\u4f60\u51d1\u5fae\u5206\u6362\u5143\u7684\u76ee\u6807\u5c31\u662f\u5c06\u88ab\u79ef\u7684\u5f0f\u5b50\u6362\u6210g[f(x)]\u4e58\u4ee5f'(x)\u7684\u5f62\u5f0f
\u7136\u540e\u5c06\u5f0f\u5b50\u6362\u6210g[f(x)]df(x) \u95ee\u9898\u5c31\u7b80\u5355\u4e86
\u222bxarccosxdx=x^2/2*arccosx+1/2\u222bx^2/\u221a(1-x^2)dx
=x^2/2*arccosx+1/2\u222bx^2/\u221a(1-x^2)dx
=x^2/2*arccosx-x/4\u221a(1-x^2)+1/4*arcsinx+C
\u222barctan\u221axdx=x*atan \u221ax-\u221ax+atan\u221ax+C
\u222barccos\u221axdx=x arccos\u221ax-1/2\u221ax\u221a(1-x) +1/2arcsin\u221ax+C
\u222barcsin\u221axdx=x arcsin\u221ax+1/2\u221ax\u221a(1-x)-1/2 arcsin\u221ax+C
∫√(1+x²)dx
=x√(1+x²)-∫xd√(1+x²)
=x√(1+x²)-∫x²/√(1+x²)dx
=x√(1+x²)-∫(x²+1-1)/√(1+x²)dx
=x√(1+x²)-∫[√(x²+1)-1/√(1+x²)]dx
=x√(1+x²)-∫√(x²+1)dx+∫1/√(1+x²)dx
移相得:
2∫√(1+x²)dx=x√(1+x²)+∫1/√(1+x²)dx
∫1/√(1+x²)dx
令x=tant dx=sec²tdt
∫1/√(1+x²)dx
=∫sec²t/sectdt
=∫secdx=ln|sect+tant|+C
=ln[x+√(1+x²)]+C
则
2∫√(1+x²)dx=x√(1+x²)+ln[x+√(1+x²)]+C
∫√(1+x²)dx=(1/2){x/√(1+x²)+ln[x+√(1+x²)]}+C
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教你一招，用MATLAB软件编程求解：
syms x
int(sqrt(1+x^2),x)
运行，
结果为1/2*x*(1+x^2)^(1/2)+1/2*asinh(x)。
其中asinh(x)是反双曲正弦ln(x+√(1+x²))。
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