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∫√(1+x)/xDx

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∫√(1+x^2) /x dx let x= tanu dx=(secu)^2 du ∫√(1+x^2) /x dx =∫(secu)^3/tanu du =∫cscu dtanu =cscu .tanu + ∫cscu du =cscu .tanu + ln|cscu-cotu| + C =√(1+x^2) + ln| √(1+x^2)/x - 1/x | + C

解 原式=Inx 希望对你有帮助 不懂追问

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

∫[√x/(1+x)] dx let y=√x dy = dx/(2√x) dx= 2ydy ∫[√x/(1+x)] dx =2∫ [y^2/(1+y^2)] dx =2∫ [ 1- 1/(1+y^2) ] dx =2( y - arctany ) + C =2( √x - arctan√x ) + C

∫(1+x)/1+xdx =∫(1+x)/1+xdx =∫1dx =x 带入上线限,有 =π/2-(-π/2) =π

令√(1+x)=t,x=t^2-1,dx=2tdt, ∫√(1+x)/xdx=2∫t^2/(t^2-1)dt =2∫(x^2-1+1)/(x^2-1)dx =2x+∫2/(x^2-1)dx =2x+∫[1/(x-1)-1/(x+1)]dx =2x+ln|(x-1)/(x+1)|+C

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