求积分∫(arctan(1/x)/(1+x^2))dx
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求积分∫(arctan(1/x)/(1+x^2))dx
求积分∫(arctan(1/x)/(1+x^2))dx
求积分∫(arctan(1/x)/(1+x^2))dx
嘿嘿,其实这题很简单.
令y = 1/x、x = 1/y、dx = - 1/y² dy
∫ [arctan(1/x)]/(1 + x²) dx
= ∫ arctany/(1 + 1/y²) * (- 1/y² dy)
= ∫ arctany * y²/(1 + y²) * (- 1/y²) dy
= - ∫ arctany/(1 + y²) dy
= - ∫ arctany d(arctany)
= (- 1/2)(arctany)² + C
= (- 1/2)[arctan(1/x)]² + C
∫arctan(1/x)/(1+x^2)*dx
=∫arctan(1/x)*d(arctanx)
=arctan(1/x)*arctanx-∫arctanx*1/[1+(1/x)^2]*(-1/x^2)dx
=arctan(1/x)*arctanx+∫arctanx*1/{[1+(1/x)^2]*x^2}*dx
=arctan(1/x)*arctanx+∫arctanx*1/(x^2+1)*dx
=arctan(1/x)*arctanx+∫arctanx*d(arctanx)
=arctan(1/x)*arctanx+(1/2)(arctanx)^2+C
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