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用倒代换 ， 令 x = 1/t, 则 dx = -dt/t^2, 得
I = ∫dx/[x√(1+x^4) = ∫-tdt/√(1+t^4) = -(1/2)∫d(t^2)/√(1+t^4)
令 t^2 = tanu, 则 dt^2 = (secu)^2du, 得
I = -(1/2)∫secudu = -(1/2)ln|secu+tanu| + C
= -(1/2)ln|√(1+t^4)+t^2| + C= -(1/2)ln|√(1+1/x^4)+1/x^2| + C
= -(1/2)ln[√(1+x^4)+1] + ln|x| + C
I = ∫dx/[x√(1+x^4) = ∫-tdt/√(1+t^4) = -(1/2)∫d(t^2)/√(1+t^4)
令 t^2 = tanu, 则 dt^2 = (secu)^2du, 得
I = -(1/2)∫secudu = -(1/2)ln|secu+tanu| + C
= -(1/2)ln|√(1+t^4)+t^2| + C= -(1/2)ln|√(1+1/x^4)+1/x^2| + C
= -(1/2)ln[√(1+x^4)+1] + ln|x| + C
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