作业帮求不定积分∫x-1/x^2+2x+3 dx 同济六版208页43题分母配方后，如果令x+1=根号2 tan t 用换元法步骤是什么
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求不定积分∫x-1/x^2+2x+3 dx 同济六版208页43题分母配方后，如果令x+1=根号2 tan t 用换元法步骤是什么
求不定积分∫x-1/x^2+2x+3 dx 同济六版208页43题
分母配方后，如果令x+1=根号2 tan t 用换元法步骤是什么
求不定积分∫x-1/x^2+2x+3 dx 同济六版208页43题分母配方后，如果令x+1=根号2 tan t 用换元法步骤是什么
原式=∫(x+1-2)/(x^2+2x+3) dx
=∫(x+1)dx/(x^2+2x+3)-∫2dx/[(x+1)^2+2]
=1/2*∫d(x^2+2x+3)/(x^2+2x+3)-2∫d(x+1)/[(x+1)^2+2]
=1/2*ln(x^2+2x+3)-根号2*[arctan(x+1)/根号2]+C
=ln[根号(x^2+2x+3)]-根号2*[arctan(x+1)/根号2]+C
∫(x-1)/[(x+1)^2+2]dx
=1/2 ∫1/[(x+1)^2+2]d(x^2+2x+3)-∫2/[(x+1)^2+2]dx
=1/2ln(x^2+2x+3)-∫2/[(x+1)^2+2]dx
=1/2ln(x^2+2x+3)-∫2/[((x+1)/根号2)^2+1]d(x+1)/根号2
=1/2ln(x^2+2x+3)-2/根号2*arctan[(x+1)/根号2]+C(常数)
=1/2ln(x^2+2x+3)-根号2*arctan[(x+1)/根号2]+C(常数)
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∫(x-1)/x^2+2x+3 dx
=∫(x-1)/[(x+1)²+2]dx
=∫x/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx
=∫(x+1-1)/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx
=∫(x+1)/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx...
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∫(x-1)/x^2+2x+3 dx
=∫(x-1)/[(x+1)²+2]dx
=∫x/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx
=∫(x+1-1)/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx
=∫(x+1)/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx-∫1/[(x+1)²+2]dx
=∫(x+1)/[(x+1)²+2]dx-2∫1/[(x+1)²+2]dx
=1/2∫1/[(x+1)²+2]d[(x+1)²+2]-2∫1/[(x+1)²+2]d(x+1)
=1/2ln[(x+1)²+2]-√2arctan(x+1)/√2+C
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∫(x-1)/(x²+2x+3)dx
=½∫(2x-2)/(x²+2x+3)dx
=½∫(2x+2-4)/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ...
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∫(x-1)/(x²+2x+3)dx
=½∫(2x-2)/(x²+2x+3)dx
=½∫(2x+2-4)/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - ½∫4/(x²+2x+3)dx
=½∫(2x+2)/(x²+2x+3)dx - 2∫1/(x²+2x+3)dx
=½∫d(x²+2x+3)/(x²+2x+3) - 2∫1/[(x+1)²+2]dx
=½ln|x²+2x+3| - ∫1/{[(x+1)/√2]²+1}dx + C
=½ln|x²+2x+3| - (√2)∫1/{[(x+1)/√2]²+1}d[(x+1)/√2] + C
=½ln|x²+2x+3| - (√2)arctan[(x+1)/√2] + C
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