不定积分∫1/{(1+x^2)(1-x^2)^(1/2)}dx

2个回答

  • Possible intermediate steps:

    integral 1/(sqrt(1-x^2) (1+x^2)) dx

    For the integrand,1/(sqrt(1-x^2) (x^2+1)) substitute x = sin(u) and dx = cos(u) du.Then sqrt(1-x^2) = sqrt(1-sin^2(u)) = cos(u) and u = sin^(-1)(x):

    = integral 1/(sin^2(u)+1) du

    For the integrand 1/(sin^2(u)+1),substitute s = tan(u/2) and ds = 1/2 sec^2(u/2) du.Then transform the integrand using the substitutions sin(u) = (2 s)/(s^2+1),cos(u) = (1-s^2)/(s^2+1) and du = (2 ds)/(s^2+1):

    = integral 2/((s^2+1) ((4 s^2)/(s^2+1)^2+1)) ds

    Simplify the integrand 2/((s^2+1) ((4 s^2)/(s^2+1)^2+1)) to get (2 (s^2+1))/(s^4+6 s^2+1):

    = integral (2 (s^2+1))/(s^4+6 s^2+1) ds

    Factor out constants:

    = 2 integral (s^2+1)/(s^4+6 s^2+1) ds

    For the integrand (s^2+1)/(s^4+6 s^2+1),factor the denominator into quadratic irreducible terms:

    = 2 integral (s^2+1)/((s^2-2 sqrt(2)+3) (s^2+2 sqrt(2)+3)) ds

    For the integrand (s^2+1)/((s^2-2 sqrt(2)+3) (s^2+2 sqrt(2)+3)),use partial fractions:

    = 2 integral ((1+sqrt(2))/(2 sqrt(2) (s^2+2 sqrt(2)+3))+(1-sqrt(2))/(2 sqrt(2) (-s^2+2 sqrt(2)-3))) ds

    Integrate the sum term by term and factor out constants:

    = 1/sqrt(2)-1 integral 1/(-s^2+2 sqrt(2)-3) ds+1+1/sqrt(2) integral 1/(s^2+2 sqrt(2)+3) ds

    The integral of 1/(-s^2+2 sqrt(2)-3) is -(tan^(-1)(s/sqrt(3-2 sqrt(2))))/sqrt(3-2 sqrt(2)):

    = 1+1/sqrt(2) integral 1/(s^2+2 sqrt(2)+3) ds-(tan^(-1)(s/sqrt(3-2 sqrt(2))))/sqrt(2 (3-2 sqrt(2)))+(tan^(-1)(s/sqrt(3-2 sqrt(2))))/sqrt(3-2 sqrt(2))

    The integral of 1/(s^2+2 sqrt(2)+3) is (tan^(-1)(s/sqrt(3+2 sqrt(2))))/sqrt(3+2 sqrt(2)):

    = -(tan^(-1)(s/sqrt(3-2 sqrt(2))))/sqrt(2 (3-2 sqrt(2)))+(tan^(-1)(s/sqrt(3-2 sqrt(2))))/sqrt(3-2 sqrt(2))+(tan^(-1)(s/sqrt(3+2 sqrt(2))))/sqrt(2 (3+2 sqrt(2)))+(tan^(-1)(s/sqrt(3+2 sqrt(2))))/sqrt(3+2 sqrt(2))+constant

    Substitute back for s = tan(u/2):

    = (tan^(-1)((tan(u/2))/sqrt(3-2 sqrt(2))))/sqrt(3-2 sqrt(2))-(tan^(-1)((tan(u/2))/sqrt(3-2 sqrt(2))))/sqrt(6-4 sqrt(2))+(tan^(-1)((tan(u/2))/sqrt(3+2 sqrt(2))))/sqrt(6+4 sqrt(2))+(tan^(-1)((tan(u/2))/sqrt(3+2 sqrt(2))))/sqrt(3+2 sqrt(2))+constant

    Substitute back for u = sin^(-1)(x):

    = (tan^(-1)(x/(sqrt(3-2 sqrt(2)) (sqrt(1-x^2)+1))))/sqrt(3-2 sqrt(2))-(tan^(-1)(x/(sqrt(3-2 sqrt(2)) (sqrt(1-x^2)+1))))/sqrt(6-4 sqrt(2))+(tan^(-1)(x/(sqrt(3+2 sqrt(2)) (sqrt(1-x^2)+1))))/sqrt(6+4 sqrt(2))+(tan^(-1)(x/(sqrt(3+2 sqrt(2)) (sqrt(1-x^2)+1))))/sqrt(3+2 sqrt(2))+constant

    Factor the answer a different way:

    = 1/2 ((2+2 sqrt(2)-sqrt(6+4 sqrt(2))) tan^(-1)(x/(sqrt(3-2 sqrt(2)) (sqrt(1-x^2)+1)))+(-2+2 sqrt(2)+sqrt(6-4 sqrt(2))) tan^(-1)(x/(sqrt(3+2 sqrt(2)) (sqrt(1-x^2)+1))))+constant

    Which is equivalent for restricted x values to:

    = (tan^(-1)((sqrt(2) x)/sqrt(1-x^2)))/sqrt(2)+constant