求不定积分,用换元法! 不定积分用换元法求解
\u6c42\u4e0d\u5b9a\u79ef\u5206\uff0c\u7528\u6362\u5143\u6cd5\u4ee4\u221a(1+t)=u\uff0c\u5f97t=u²-1\uff0cdt=2udu
\u222b1/[1+\u221a(1+t)]dt
=\u222b2u/(1+u)du
=2\u222b[(1+u)-1]/(1+u)du
=2\u222bdu-2\u222b1/(1+u)d(1+u)
=2u-2ln(1+u)+C
=2\u221a(1+t)-2ln[1+\u221a(1+t)]+C
\u4ee4\u221a(x²+a²)=t\uff0c\u5f97x²=t²-a²\uff0cdx²=2tdt
\u222b\u221a(x²+a²)/xdx
=\u222bx\u221a(x²+a²)/x²dx
=[\u222b\u221a(x²+a²)/x²dx²]/2
=[\u222b2t•t/(t²-a²)dt]/2
=\u222b[(t²-a²)+a²]/(t²-a²)dt
=\u222bdt+a²\u222b1/(t²-a²)dt
=t+aln[(t-a)/(t+a)]/2+C
=\u221a(x²+a²)+aln{[\u221a(x²+a²)-a]/[\u221a(x²+a²)+a]}/2+C
\u4ee4\u221a(1+2/x)=u\uff0c\u5f97x=2/(u²-1)\uff0cdx=-4u/(u²-1)²
\u222b\u221a(x²+2x)/x²dx
=\u222b\u221a[4/(u²-1)²+4/(u²-1)]/[4/(u²-1)²]•[-4u/(u²-1)²]du
=-\u222bu\u221a[4+4(u²-1)/(u²-1)²]du
=-2\u222bu²/(u²-1)du
=-2\u222b[(u²-1)+1]/(u²-1)du
=-2\u222bdu-2\u222b1/(u²-1)du
=ln[(1+u)/(1-u)]-2u+C
=ln[(1+\u221a(1+2/x))/(1-\u221a(1+2/x))]-2\u221a(1+2/x)+C
\u6b64\u5904\u221a(1+2/x)=u\u7ecf\u8fc7\u4e24\u6b21\u4ee3\u6362
\u9996\u5148\u4ee4x=1/t\uff0c\u5f97dx=-1/t²dt\uff0c\u5f97\u5230\u222b\u221a(1+2t)/tdt\uff0c\u518d\u4ee4\u221a(1+2t)=u\uff0c\u5373\u221a(1+2/x)=u
\u4ee4\u221a(e^u+1)=t\uff0c\u5f97u=ln(t²-1)\uff0cdu=2t/(t²-1)dt
\u222b1/\u221a(e^u+1)du
=\u222b1/t•2t/(t²-1)dt
=\u222b1/(t²-1)dt
=ln[(t-1)/(t+1)]+C
=ln[(\u221a(e^u+1)-1)/(\u221a(e^u+1)+1)]+C
\u4ee4x=1/t\uff0c\u5f97dx=-1/t²dt
\u222b1/x\u221a(a²-b²x²)dx
=-\u222bt/\u221a(a²-b²/t²)•1/t²dt
=-\u222bt²/\u221a(a²t²-b²)•1/t²dt
=-\u222b1/a\u221a[t²-(b/a)²]dt
=-ln[t+\u221a(t²-b²/a²)]/a+C
=-ln[1/x+\u221a(1/x²-b²/a²)]/a+C
=ln{ax/[a+\u221a(a²-b²x²)]}/a+C
\u4ee4\u221a(1+lnx)=t\uff0c\u5f97x=e^(t²-1)\uff0cdx=2te^(t²-1)
\u222b\u221a(1+lnx)/xlnxdx
=\u222bt/(t²-1)e^(t²-1)•2te^(t²-1)dt
=2\u222bt²/(t²-1)dt
=2\u222b[(t²-1)+1]/(t²-1)dt
=2\u222bdt+2\u222b1/(t²-1)dt
=2t+ln[(t-1)/(t+1)]+C
=2\u221a(1+lnx)+ln[(\u221a(1+lnx)-1)/(\u221a(1+lnx)+1)]+C
\u8bbe x\uff1du² - 1\uff0c\u5219 dx\uff1d2udu\uff0c
\u539f\u5f0f\uff1d\u222b2du / (u² - 1)²
\uff1d1/2 ln|u\uff0b1| - 1/2 ln|u - 1| - u/(u²-1)\uff0bC
\uff1d(\u81ea\u5df1\u56de\u4ee3)
∫1/√(x^2+1)^3 dx
=∫1/sec³t * (sec²t dt)
=∫cost dt
= sint + C
= tant*cost + C
= x/√(x^2+1) + C
2）令: x=t^6 ,
∫1/[√x + ³√x ] dx
=∫1/[t² + t³] (6t^5 dt)
= 6*∫t^3/[1 + t] dt
= 6*∫[(t^3+1)-1]/[1 + t] dt
= 6*∫[(t^2 - t +1) -1/(1+t) ]dt
= 2*t^3 - 3*t^2 + 6*t -6*ln(1+t) + C
= 2*√x - 3*³√x + 6*x^(1/6) - 6*ln(1+x^(1/6)) + C
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