∫ln[x+(1+x)^(1/2)]dx这个怎么做?∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
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∫ln[x+(1+x)^(1/2)]dx这个怎么做?∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
∫ln[x+(1+x)^(1/2)]dx这个怎么做?∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
∫ln[x+(1+x)^(1/2)]dx这个怎么做?∫[(3x+1)/(4+x^2)^(1/2)]dx高等数学,
1.设√(1+x)= u 则 x= u^2 -1
原式 = ln(u^2 - 1 + u) d(u^2 -1) 然后分部积分做
2.x = 2tan t
∫ln[x+(1+x^2)^(1/2)]dx=∫(x)'*ln[x+(1+x^2)^(1/2)]dx
=x*ln[x+(1+x^2)^(1/2)]—∫x*(1+x^2)^(-1/2)dx
=x*ln[x+(1+x^2)^(1/2)]+(1+x^2)^(1/2)+C
∫[(3x+1)/(4+x^2)^(1/2)]dx
令(4+x^2)^(1/2)=t,所以x=(t^2...
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∫ln[x+(1+x^2)^(1/2)]dx=∫(x)'*ln[x+(1+x^2)^(1/2)]dx
=x*ln[x+(1+x^2)^(1/2)]—∫x*(1+x^2)^(-1/2)dx
=x*ln[x+(1+x^2)^(1/2)]+(1+x^2)^(1/2)+C
∫[(3x+1)/(4+x^2)^(1/2)]dx
令(4+x^2)^(1/2)=t,所以x=(t^2—4)^(1/2),dx=t/(t^2—4)^(1/2)dt,
所以原式=∫[3(t^2—4)^(1/2)+1]/t*t/(t^2—4)^(1/2)dt=3∫dt+∫1/(t^2—4)^(1/2)dt
=3t+㏑[t+(t^2—4)^(1/2)]+C
=3(4+x^2)^(1/2)+㏑[(4+x^2)^(1/2)+x]+C
收起
令x=tanx,试一下应该可以解决的