Question

如图，求不定积分 要详细过程

Answer 1

let
x^2 = sinu
2x dx = cosu du
∫x.√[ (1-x^2)/(1+x^2) ] dx
=∫x.(1-x^2)/√(1-x^4) dx
= ∫x/√(1-x^4) dx - ∫x^3/√(1-x^4) dx
= ∫x/√(1-x^4) dx + (1/4) ∫d(1-x^4)/√(1-x^4)
= ∫x/√(1-x^4) dx + (1/2) √(1-x^4)
= (1/2) ∫ cosu du/cosu + (1/2) √(1-x^4)
= (1/2) ∫ du + (1/2) √(1-x^4)
= (1/2) u + (1/2) √(1-x^4) +C
= (1/2) arcsin(x^2) + (1/2) √(1-x^4) +C

Answer 2

们推荐一下呗

Answer 3

令√[(1-x²)/(1+x²)]=t，则x²=(1-t²)/(1+t²)
∫x√[(1-x²)/(1+x²)]dx

=½∫√[(1-x²)/(1+x²)]d(x²)
=½∫td[(1-t²)/(1+t²)]
=½t(1-t²)/(1+t²) - ½∫[(1-t²)/(1+t²)]dt
=½t(1-t²)/(1+t²) - ∫[1/(1+t²) -½]dt
=½t(1-t²)/(1+t²) - arctant+½t+C
=½x²√[(1-x²)/(1+x²)] -arctan[√[(1-x²)/(1+x²)]] +½√[(1-x²)/(1+x²)] +C