Question

一道高数题：求不定积分∫(x^2+x^4)^1/2dx.

Answer 1

∫ (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

Answer 2

∫ √(x² + x⁴) dx
= ∫ |x|√(1 + x²) dx
= (1/2)∫ √(1 + x²) d(x² + 1)
= (1/2)(2/3)(1 + x²)^(3/2) + C
= (1/3)(1 + x²)^(3/2) + C