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∫√1+x/1-xdx
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∫√1+x/1-xdx
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答案和解析
令y=√(1+x),则x=y^2-1
原式=∫y/(2-y^2)d(y^2-1)
=2∫y^2/(2-y^2)dy
=-2∫[1+2/(y^2-2)]dy
=-2∫dy-√2∫[1/(y-√2)-1/(y+√2)]dy
=-2y-√2[ln(y-√2)-ln(y+√2)]+C
=-2y-√2ln[(y-√2)/(y+√2)]+C
=-2√(1+x)-√2ln[(√(1+x)-√2)/(√(1+x)+√2)]+C
原式=∫y/(2-y^2)d(y^2-1)
=2∫y^2/(2-y^2)dy
=-2∫[1+2/(y^2-2)]dy
=-2∫dy-√2∫[1/(y-√2)-1/(y+√2)]dy
=-2y-√2[ln(y-√2)-ln(y+√2)]+C
=-2y-√2ln[(y-√2)/(y+√2)]+C
=-2√(1+x)-√2ln[(√(1+x)-√2)/(√(1+x)+√2)]+C
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