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1-x-1-x-dx-




Question Number 492 by 123456 last updated on 25/Jan/15
∫((1−(√x))/(1+(√x)))dx
1x1+xdx
Answered by prakash jain last updated on 14/Jan/15
x=t^2   dx=2t dt  ∫((1−t)/(1+t))∙2t dt=2∫ ((t−t^2 )/(1+t)) dt  =2∫((1+t+1−t^2 −2)/(1+t))  =2[∫1 dt+∫(1−t)dt−∫(2/(1+t)) dt]  =2[t+t−(t^2 /2)−2ln ∣1+t∣]+C  =4t−t^2 −4ln ∣1+t∣+C  =4(√x)−x−4ln ∣1+(√x)∣+C
x=t2dx=2tdt1t1+t2tdt=2tt21+tdt=21+t+1t221+t=2[1dt+(1t)dt21+tdt]=2[t+tt222ln1+t]+C=4tt24ln1+t+C=4xx4ln1+x+C

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