重点在∫x²·1/(x+1)dx如何转换到下一步答案是1/2[(x²+1)ln(x²+1)-x²]+C

重点在∫x²·1/(x+1)dx如何转换到下一步答案是1/2[(x²+1)ln(x²+1)-x²]+C
重点在∫x²·1/(x+1)dx如何转换到下一步
答案是1/2[(x²+1)ln(x²+1)-x²]+C

重点在∫x²·1/(x+1)dx如何转换到下一步答案是1/2[(x²+1)ln(x²+1)-x²]+C
∫xln(x²+1)dx
=(1/2)∫ln(x²+1)dx²
=(1/2)[x²ln(x²+1)-(x²-ln(x²+1))]+C
=(1/2)(X²+1)ln(x²+1)-(x²/2)+C

分部积分
∫ln(x^2+1)dx = ∫x d ln(x^2+1) = xln(x^2+1) - ∫x d ln(x^2+1)
= xln(x^2+1) - 2∫(x^2/x^2+1)dx
= xln(x^2+1) - 2∫(x^2+1-1)/(x^2+1)dx
= xln(x^2+1) - 2[∫(x^2+1)/(x^2+1)dx -∫(1/x^2+1)dx]...

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分部积分
∫ln(x^2+1)dx = ∫x d ln(x^2+1) = xln(x^2+1) - ∫x d ln(x^2+1)
= xln(x^2+1) - 2∫(x^2/x^2+1)dx
= xln(x^2+1) - 2∫(x^2+1-1)/(x^2+1)dx
= xln(x^2+1) - 2[∫(x^2+1)/(x^2+1)dx -∫(1/x^2+1)dx]
= xln(x^2+1) - 2[∫dx -∫(1/x^2+1)dx]
= xln(x^2+1) - 2[x - arctanx]+C

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