一道高数题:求不定积分∫(x^2+x^4)^1/2dx.
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一道高数题:求不定积分∫(x^2+x^4)^1/2dx.
![一道高数题:求不定积分∫(x^2+x^4)^1/2dx.](/uploads/image/z/6843973-13-3.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E9%A2%98%EF%BC%9A%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%28x%5E2%2Bx%5E4%29%5E1%2F2dx.)
∫ (x^2+x^4)^1/2 dx
= ∫[x^2(x^2+x^4)]^1/2dx
= ∫x[(1+x^2)]^1/2dx 设x=tant,则dx=(sect)^2dt ∫(x^2+x^4)^1/2dx
= ∫[x^2(x^2+x^4)]^1/2dx
= ∫x[(1+x^2)]^1/2dx
= ∫tant{[1+(tant)^2]}^1/2* (sect)^2dt
= ∫tantsect* (sect)^2dt
= ∫(sect)^2d(secttant)
= (1/3)(sect)^3+C
由tant=x得到,sect=(x^2+1)^(1/2)
∫(x^2+x^4)^1/2dx =(1/3)(x^2+1)^(3/2)+C
= ∫[x^2(x^2+x^4)]^1/2dx
= ∫x[(1+x^2)]^1/2dx 设x=tant,则dx=(sect)^2dt ∫(x^2+x^4)^1/2dx
= ∫[x^2(x^2+x^4)]^1/2dx
= ∫x[(1+x^2)]^1/2dx
= ∫tant{[1+(tant)^2]}^1/2* (sect)^2dt
= ∫tantsect* (sect)^2dt
= ∫(sect)^2d(secttant)
= (1/3)(sect)^3+C
由tant=x得到,sect=(x^2+1)^(1/2)
∫(x^2+x^4)^1/2dx =(1/3)(x^2+1)^(3/2)+C