Question

求不定积分∫(1+x^2)^1/2dx

Answer 1

你好！可以拆成两项如图，都是简单的积分。经济数学团队帮你解答，请及时采纳。谢谢！

Answer 2

不必换元，可直接计算。如图

Answer 3

令x=tant,t∈(-π/2,π/2),则√(1+x²)=sect,dx=sec²tdt
∫√(1+x²) dx
=∫sec³t dt
=∫sect d(tant)
=sect*tant-∫tant d(sect)
=sect*tant-∫tan²t*sectdt
=sect*tant-∫(sec²t-1)*sectdt
=sect*tant-∫sec³tdt+∫sectdt
∴∫sec^3tdt=(1/2)(sect*tant+∫sectdt)
=(1/2)(sect*tant+ln|sect+tant|)+C

Answer 4

∫x^2/(1+x^2)^2 dx =-(1/2)∫xd(1/(1+x^2)) =-(1/2)[x/(1+x^2)] + (1/2)∫ dx/(1+x^2) =-(1/2)[x/(1+x^2)] + (1/2)arctanx + C

Answer 5

∫(x+1/x)² dx 
= ∫(x² + 2 + 1/x²) dx 
= ∫x²dx + ∫2dx + ∫1/x² dx 
= x ³ / 3 + 2x + (-1 / x) + C
= x ³ / 3 + 2x - 1 / x + C