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求定积分上限是5下限是0(x^3/x^2+1)dx麻烦写一下过程
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求定积分上限是5下限是0 (x^3/x^2+1)dx麻烦写一下过程
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答案和解析
x³/(x²+1)
=(x³+x-x)/(x²+1)
=x-x/(x²+1)
所以原式=∫(0,5)xdx-∫(0,5)x/(x²+1)dx
=∫(0,5)xdx-1/2∫(0,5)d(x²+1)/(x²+1)
=x²/2-1/2*ln(x²+1) (0,5)
=25/2-ln√26
=(x³+x-x)/(x²+1)
=x-x/(x²+1)
所以原式=∫(0,5)xdx-∫(0,5)x/(x²+1)dx
=∫(0,5)xdx-1/2∫(0,5)d(x²+1)/(x²+1)
=x²/2-1/2*ln(x²+1) (0,5)
=25/2-ln√26
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