∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?把式子因数化,变成(?)/(x^2+1)+(?)/(x-2)步,带骤,

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∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?把式子因数化,变成(?)/(x^2+1)+(?)/(x-2)步,带骤,
∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?
把式子因数化,变成(?)/(x^2+1)+(?)/(x-2)步,带骤,

∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?把式子因数化,变成(?)/(x^2+1)+(?)/(x-2)步,带骤,
∵(x^4-4x^2+5x-15)/[(x^2+1)(x-2)]
=[(x^4+x²-5x²-5)+(5x-10)]/[(x²+1)(x-2)]
=[x²(x²+1)-5(x²+1)+5(x-2)]/[(x²+1)(x-2)]
=[(x²+1)(x²-5)+5(x-2)]/[(x²+1)(x-2)]
=(x²-5)/(x-2)+5/(x²+1)
=[(x²-2x)+(2x-4)-1]/(x-2)+5/(x²+1)
=[x(x-2)+2(x-2)-1]/(x-2)+5/(x²+1)
=x+2-1/(x-2)+5/(x²+1)
∴∫(x^4-4x^2+5x-15)/(x^2+1)(x-2)dx
=∫[x+2-1/(x-2)+5/(x²+1)]dx
=x²/2+2x-ln|x-2|+5arctanx+C,(C是积分常数).

分解为 (x^3+x+5)/(x^2+1)+(2x-5)/(x-2)=5/(x^2+1)-1/(x-2)+x+2
再求积分得5arctanx-log(x-2)+x^2/2+2x+一个常数

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