2．求不定积分∫1nxdx 求1/1+tanx的不定积分
\u6c42\u4e0d\u5b9a\u79ef\u5206\u222b1/\uff082+cosx\uff09dx\u222b1/\uff082+cosx\uff09dx=2/\u221a3arctan[tan(x/2)/\u221a3]+C\u3002C\u4e3a\u5e38\u6570\u3002
\u89e3\u7b54\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u8bbet=tan(x/2)
\u5219cosx=[cos²(x/2)-sin²(x/2)]/[cos²(x/2)+sin²(x/2)]
=[1-tan²(x/2)]/[1+tan²(x/2)]
=(1-t²)/(1+t²)
dx=d(2arctant)=2dt/(1+t²)
\u6545\u222b1/(2+cosx)dx=\u222b1/[2+(1-t²)/(1+t²)]*[2dt/(1+t²)]
=\u222b2dt/(3+t²)
=2/\u221a3\u222bd(t/\u221a3)/[1+(t/\u221a3)²]
=2/\u221a3arctan(t/\u221a3)+C
=2/\u221a3arctan[tan(x/2)/\u221a3]+C
\u6269\u5c55\u8d44\u6599\uff1a
\u5206\u90e8\u79ef\u5206\uff1a
(uv)'=u'v+uv'
\u5f97\uff1au'v=(uv)'-uv'
\u4e24\u8fb9\u79ef\u5206\u5f97\uff1a\u222b u'v dx=\u222b (uv)' dx - \u222b uv' dx
\u5373\uff1a\u222b u'v dx = uv - \u222b uv' d,\u8fd9\u5c31\u662f\u5206\u90e8\u79ef\u5206\u516c\u5f0f
\u4e5f\u53ef\u7b80\u5199\u4e3a\uff1a\u222b v du = uv - \u222b u dv
\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
7\uff09\u222bcosxdx=sinx+c
8\uff09\u222b1/(cosx)^2dx=tanx+c
9\uff09\u222b1/(sinx)^2dx=-cotx+c
10\uff09\u222b1/\u221a\uff081-x^2) dx=arcsinx+c
\u222b1/tanx dx
=\u222bcosx/sinx dx
=\u222b1/sinx dsinx
=ln|sinx|+C
\u6269\u5c55\u8d44\u6599\uff1a
\u7531\u4e8e\u5728\u4e00\u4e2a\u533a\u95f4\u4e0a\u5bfc\u6570\u6052\u4e3a\u96f6\u7684\u51fd\u6570\u5fc5\u4e3a\u5e38\u6570\uff0c\u6240\u4ee5G(x)-F(x)=C\u2019(C\u2018\u4e3a\u67d0\u4e2a\u5e38\u6570)\u3002
\u8fd9\u8868\u660eG(x)\u4e0eF(x)\u53ea\u5dee\u4e00\u4e2a\u5e38\u6570.\u56e0\u6b64,\u5f53C\u4e3a\u4efb\u610f\u5e38\u6570\u65f6\uff0c\u8868\u8fbe\u5f0fF(x)+C\u5c31\u53ef\u4ee5\u8868\u793af(x)\u7684\u4efb\u610f\u4e00\u4e2a\u539f\u51fd\u6570\u3002\u4e5f\u5c31\u662f\u8bf4f(x)\u7684\u5168\u4f53\u539f\u51fd\u6570\u6240\u7ec4\u6210\u7684\u96c6\u5408\u5c31\u662f\u51fd\u6570\u65cf{F(x)+C|-\u221e<C<+\u221e}\u3002
\u7531\u6b64\u53ef\u77e5\uff0c\u5982\u679cF(x)\u662ff(x)\u5728\u533a\u95f4I\u4e0a\u7684\u4e00\u4e2a\u539f\u51fd\u6570\uff0c\u90a3\u4e48F(x)+C\u5c31\u662ff(x)\u7684\u4e0d\u5b9a\u79ef\u5206\uff0c\u5373\u222bf(x)dx=F(x)+C\u3002
\u56e0\u800c\u4e0d\u5b9a\u79ef\u5206\u222bf(x) dx\u53ef\u4ee5\u8868\u793af(x)\u7684\u4efb\u610f\u4e00\u4e2a\u539f\u51fd\u6570\u3002
(uv)' = vu' + uv'
uv = ∫v du + ∫u dv
∫v du = uv - ∫u dv
对于∫lnx dx
可设u = x，v = lnx
du = dx，dv = d(lnx) = (1/x) dx
代入公式就是∫lnx dx = xlnx - ∫x dlnx
= xlnx - ∫[x * (1/x) dx]
= xlnx - ∫ dx
= xlnx - x + C
你这没有定积分上限和下限啊。
先求原函数，为1/x
然后带入上限减去下限
上面瞎扯，用分步积分：
∫lnxdx=xlnx-∫xdlnx=xlnx-∫x(lnx)'dx=xlnx-x+C
=x*lnx-x+C
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