求In(√(x^2+1)+x)/(1-x^2)^2的不定积分,
1个回答
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
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