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时间:2011-12-08 17:38
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dxrepair2.1
∫ xd ln(1+x^2)=?_百度知道
∫ xd ln(1+x^2)=?
提问者采纳
d ln(1+x²)=2x/(1+x²) dx因此:原式=∫ 2x²/(1+x²) dx=2∫ (x²+1-1)/(1+x²) dx=2∫ 1 dx - 2∫ 1/(1+x²) dx=2x - 2arctanx + C希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的&选为满意回答&按钮,谢谢。
提问者评价
原来是这样,感谢!
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出门在外也不愁百度知道搜索_求解微分方程dy/dx=x^2/y(1+x^3)求解 ∫dx/xxxxxxxx(1-x×x)_百度知道
求解 ∫dx/xxxxxxxx(1-x×x)
提问者采纳
,∫dx /,[x^8(1-x^2)] = a1/,2put x=-12b=1 =>,a1=01/,x^4-1/,[x^8(1-x^2)]let1/,+a8/, of x^6a2-a4=0
=>,x+a2/,a3=0coef,(7x^7) + (1/,2(1/, b=1/,(1-x)1= a1x^7(1-x)(1+x)+a2x^6(1-x)(1+x)+,,a2=-1coef, of x^2a6+a8 =0
=>,,x^4-1/,x^2-1/,x^8+ 1/,(3x^3)+1/,,(1-x),x^8+ b/,2(1/,(1-x))∫dx /, of xa7= 0coef,(1+x) + c/,2(1/,x^6+ 1/, of x^5a3-a5=0
=>,+a8(1-x)(1+x)+ b(1-x)x^8+c(1+x)x^8put x=12c=1
=>,x +1/,[x^8(1-x^2)] =-1/, c=1/,x^6+ 1/, a4=-1coef,x^2-1/,2put x=0a8=1coef,a6=-1coef, of x^7a1-a3=0
=>, of x^3a5-a7=0
=>,(1+x)) + 1/,a5=0coef,(1+x)/, + C,(5x^5) - 1/, of x^4a4-a6=0 =>,x^8+ 1/,[x^8(1-x^2)]=∫[-1/,2) ln,(1+x)) + 1/,2(1/,(1-x))] dx= 1/,x^2+,
提问者评价
神呐,谢啦
其他类似问题
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其他3条回答
2)(arcsinx+ln,2)∫d(sint+cost)/,(sint+cost)] dt=(1/,令x=sint, 则√(1-x²,2 + (1/,)=cost, dx=costdt∴原式=∫ cost/,2)(t+ln,2)∫(cost-sint)/,),) + C=(1/,x+√(1-x²,)+CC为任意常数,(sint+cost) dt=(1/,2)∫ dt + (1/,2)∫[(cost+sint)+(cost-sint)]/,(sint+cost) dt=t/,(sinx+cosx)=(1/,sint+cost,
xxxxxxxx是什么呀,是x的8次方吗?
是的,不好意思。。。
这道题挺2的
等待您来回答
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