| 部分積分により ∫xlog(x+√(1+x^2))dx =(1/2)(x^2)log(x+√(1+x^2)) -(1/2)∫(x^2){(1+x/√(1+x^2))/(x+√(1+x^2))}dx =(1/2)(x^2)log(x+√(1+x^2)) -(1/2)∫(x^2){(x+√(1+x^2))/{(x+√(1+x^2))√(1+x^2)}}dx =(1/2)(x^2)log(x+√(1+x^2))-(1/2)∫{(x^2)/√(1+x^2)}dx =(1/2)(x^2)log(x+√(1+x^2))-(1/2)∫√(1+x^2)dx+(1/2)∫dx/√(1+x^2) (A) ここで更に部分積分により ∫√(1+x^2)dx=x√(1+x^2)-∫{(x^2)/√(1+x^2)}dx =x√(1+x^2)-∫√(1+x^2)dx+∫dx/√(1+x^2) ∴∫√(1+x^2)dx=(1/2)x√(1+x^2)+(1/2)∫dx/√(1+x^2) ですので(A)は ∫xlog(x+√(1+x^2))dx =(1/2)(x^2)log(x+√(1+x^2))-(1/4)x√(1+x^2)+(1/4)∫dx/√(1+x^2) 第三項の積分で x=tanθ と置くと ∫dx/√(1+x^2) =∫(cosθ){dθ/(cosθ)^2} =∫{(cosθ)/{1-(sinθ)^2}}dθ =(1/2)∫{1/(1-sinθ)+1/(1+sinθ)}(cosθ)dθ =… ですので…。
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