 | 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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