换元求不定积分，转化不下去了，求解
\u4e3a\u4ec0\u4e48\u6c42\u4e0d\u5b9a\u79ef\u5206\u8981\u6362\u5143\u4e0d\u5b9a\u79ef\u5206\u662f\u6c42\u5bfc\u4e0e\u5fae\u5206\u7684\u9006\u8fd0\u7b97\uff0c\u8981\u7406\u89e3\u4e0d\u5b9a\u79ef\u5206\u4e2d\u7684\u6982\u5ff5\u4e0e\u65b9\u6cd5\u3002\u5e94\u8be5\u4ece\u6c42\u5bfc\u4e0e\u5fae\u5206\u90a3\u91cc\u60f3\u3002\u5fae\u5206\u6cd5\u4e2d\u6709\u4e2a\u91cd\u8981\u6027\u8d28\u53eb\u505a\u5fae\u5206\u5f62\u5f0f\u7684\u4e0d\u53d8\u6027\uff0c\u5373\uff0c\u5fae\u5206dy=f\u2018(u)du,\u8fd9\u91cc\u7684u\u5373\u53ef\u4ee5\u662f\u81ea\u53d8\u91cf\u4e5f\u53ef\u4ee5\u662f\u4e2d\u95f4\u53d8\u91cf\u3002\u4e0d\u5b9a\u79ef\u5206\u8981\u6362\u5143\u6cd5\u5bf9\u5e94\u7740\u590d\u5408\u51fd\u6570\u7684\u6c42\u5bfc\u6216\u5fae\u5206\u3002\u4f60\u6240\u8bf4\u7684\u201c\u8981\u4fdd\u8bc1\u90a3\u53d8\u91cf\u4e00\u6837\u201d\u662f\u56e0\u4e3a\u6362\u5143\u5c31\u662f\u8981\u628a\u88ab\u79ef\u51fd\u6570\u5316\u4e3a\u4e2d\u95f4\u53d8\u91cfu\u7684\u5bfc\u6570\uff0c\u5f53\u7136\u8981\u4e58\u4e2d\u95f4\u53d8\u91cf\u7684\u5fae\u5206du.
\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
sect=√（1+x^2),
cost=1/√（1+x^2)
sint=x/√(1+x^2)
原式=∫(sect)^2dt/[（tant)^2*sect]
=∫(cost)^2sectdt/(sint)^2
=∫costdt/(sint)^2
=∫d(sint)/(sint)^2
=-1/sint+C
=-[√（1+x^2)]/x+C.
你第二步都换错了。。。。。。x是变量。。。怎么能提到积分号外面去。。。。
定积分在换元的时候，上下限也会跟着变，所以不熟的话最好每一步都把上下限标出来。
这题单用换元积分法我做不出来。。。。。先令1+x^2=t^2，把根号去掉，然后再用分部积分法算。
这个题用三角换元
设 x=tank 用S代表积分符豪
原式=S[1/(tank)^2 *seck ]*(seck)^2 dk
=Scosk/(sink)^2 dk （因为 csck求导是等于-cotk *csck）
=-Sdcsck
=-csck+C
然后 把k=arctanx 带入 即得 原式=-csc[arctanx] +C
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