Question

求In(√(x^2+1)+x)/(1-x^2)^2的不定积分,

Answer 1

原式子=∫1/2ln(√(x^2+1)+x)d[1/(1-x^2)]

=1/2*[1/(1-x^2)]*ln(√(x^2+1)+x)-1/2∫[1/(1-x^2)]*1/(1+x^2)^(1/2)dx

令x=tant

∫[1/(1-x^2)]*1/(1+x^2)^(1/2)dx

=∫cost/(cost^2-sint^2)dt

=∫1/(1+(sint)^2)dt 令sint=u

=∫1/(1-2u^2)du

=∫1/(1-2^(1/2)u)(1+2^(1/2)u)du

=[2^(1/2)/4]*ln|[1+2^(1/2)u]/|[1-2^(1/2)u]|

=[2^(1/2)/4]*ln|[1+2^(1/2)sint]/|[1-2^(1/2)sint]|

=[2^(1/2)/4]*ln|[1+2^(1/2)(x/(1+x^2)^(1/2))]/|[1-2^(1/2))(x/(1+x^2)^(1/2))]|

带入原式得=

1/2*[1/(1-x^2)]*ln(√(x^2+1)+x)-2^(1/2)/8]*ln|[1+2^(1/2)*(x /√(x^2+1))]/|[1-2^(1/2)*(x /√(x^2+1))]| +C