几道不定积分题
\u6c42\u51e0\u9053\u4e0d\u5b9a\u79ef\u5206\u7684\u9898\u6df1\u591c\u3002\u3002\u3002
1\u3001\u539f\u5f0f=[1/(r+1)]\u222b{1/[\u221a[x^2(r+1)+1]]}d(x^(r+1))
=[1/(r+1)]ln|x^(r+1)+\u221a[x^(2r+2)+1]|+C
2\u3001\u539f\u5f0f=\u222b{1/\u221a[(x-1)^2+2^2]}d(x-1)
=ln|x-1+\u221a(x^2-2x+5)|+C
3\u3001\u539f\u5f0f=-\u222bxd(cscx)
=-xcscx+\u222bcscxdx
=-xcscx-cscxcotx+C
4\u3001\u4ee4t=e^x
\u539f\u5f0f=\u222b[ln(t+1)]/(t^2)d(t)
=-\u222bln(t+1)d(1/t)
=-[ln(t+1)]/t+\u222b1/[t(t+1)]dt
=-[ln(t+1)]/t+\u222b(1/t)dt-\u222b[1/(t+1)]dt
=-[ln(t+1)]/t+lnt-ln(t+1)+C
=-[ln(t+1)]/t+ln[t/(t+1)]+C
5\u3001\u4ee4t=arctanx,
x=tant dx=sec^2tdt
\u539f\u5f0f=\u222bttantsectdt
=\u222btd(sect)
=tsect-\u222bsectdt
=tsect-ln|tant+sect|+C
\u5c06t=arctanx\u4ee3\u56de\u4e0a\u5f0f\u5373\u4e3a\u7b54\u6848
6\u3001\u8fd9\u9898\u7528\u624b\u673a\u6253\u590d\u6742\uff0c\u8bf4\u8bf4\u601d\u8def\u5427\u3002
\u539f\u5f0f=(1/12)\u222b(1/(x^8+3x^4+2)d(x^12)
\u5bf9\u4e0a\u5f0f\u4ee4t=x^4
\u5316\u7b80\u5f97\uff1a(1/4)\u222b(t^2)/(t^2+3t+2)dt
\u5206\u5b50\u52a03t+2-3t-2\u7136\u540e\u62c6\u9879\uff0c\u518d\u5229\u7528\u6709\u7406\u51fd\u6570\u5206\u89e3\u56e0\u5f0f\u4e4b\u7c7b\u7684\u518d\u62c6\u9879\uff0c\u518d\u51d1\u5fae\u5206\uff0c\u518d\u5957\u5e38\u7528\u516c\u5f0f\u3002\u8bf4\u8d77\u6765\u590d\u6742\uff0c\u5176\u5b9e\u5e76\u4e0d\u590d\u6742\u3002
\u7761\u89c9\u53bb\u3002\u3002\u3002
\u90fd\u662f\u6bd4\u8f83\u57fa\u7840\u7684\u9898\u76ee\u4e86\uff0c\u53ea\u8981\u719f\u7ec3\u638c\u63e1\u6362\u5143\u79ef\u5206\u6cd5\u3001\u5206\u90e8\u79ef\u5206\u6cd5\u7b49\u57fa\u672c\u6280\u5de7\u5c31\u53ef\u4ee5\u4e86\uff0c
\u8fc7\u7a0b\u89c1\u56fe\u7247\uff1a\uff08\u4e24\u9053\u4e00\u6837\u7684\u95ee\u9898\uff0c\u90fd\u7ed9\u5206\u6211\u5427\uff0cword\u6253\u516c\u5f0f\u4e0d\u5bb9\u6613......\u8c22\u8c22\uff01\uff09
1、∫sin(x/3) dx＝3∫sin(x/3) d(x/3)＝-3cos(x/3)＋C 2、∫cos^3x dx＝∫(cosx)^2dsinx＝∫[1-(sinx)^2]dsinx＝sinx-1/3×(sinx)^3＋C3、∫(sin√x/)(√x) dx＝2∫(sin√x/)d(√x)＝-2cos(√x)＋C4、∫arctanx/1+x^2 dx＝∫arctanxdarctanx＝1/2×(arctanx)^2＋C
1、∫sin(x/3) dx =-3∫ d(cosx/3)=-3cos(x/3)+c
2、∫cos^3x dx=∫cos²x*cosx dx=∫(1-sin²x)d(sinx)=sinx-sin³x/3+c
3、∫(sin√x/)(√x) dx?????
4、∫arctanx/(1+x^2) dx=∫arctanx d(arctanx)=(arctanx)²/2+c
第3题的题目/后面是什么？？
希望对你有所帮助
如有问题，可以追问。
谢谢采纳
1、原式=3∫sin(x/3)d(x/3)=-3cos(x/3)+C
2、原式=∫cosx(1-sin^2x)dx=∫(1-sin^2x)d(sinx)=sinx-(sin^3x)/3+C
3、原式=2∫(sin√x)d(√x)=-2cos(√x)+C
4、令x=tant dx=sec^2tdt
原式=∫arctan(tant)/sec^2t*sec^2tdt=∫tdt=[(arctanx)^2]/2+C
1、∫sin(x/3) dx＝3∫sin(x/3) d(x/3)＝-3cos(x/3)＋C
2、∫cos^3x dx＝∫(cosx)^2dsinx＝∫[1-(sinx)^2]dsinx＝sinx-1/3×(sinx)^3＋C
3、∫(sin√x/)(√x) dx＝2∫(sin√x/)d(√x)＝-2cos(√x)＋C
4、∫arctanx/1+x^2 dx＝∫arctanxdarctanx＝1/2×(arctanx)^2＋C
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