InterviewSolution
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InterviewSolution
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Evaluate : (i) `int(dx)/(1+sqrt(x))` (ii) `int(x+sqrt(x+1))/(x+2)dx` |
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Answer» (i) Put `sqrt(x)=t` so that `x=t^(2)anddx=2tdt`. `:.int(dx)/(1+sqrt(x))=int(2t)/((1+t))dt=int(2(1+t)-2)/((1+t))dt` `=2intdt-2int(dt)/((1+t))=2t-2log|1+t|+C` `=2sqrt(x)-2log|1+sqrt(x)|+C`. (ii) Put `sqrt(x+1)=t` so that `x+1=t^(2) and dx=2tdt`. `:.int(x+sqrt(x+1))/((x+2))dx=2int((t^(2)-1+t)t)/((t^(2)+1))dt` `=2int((t^(3)+t^(2)-t)/(t^(2)+1))dt` `=2int(t+1-(2t+1)/(t^(2)+1))dt" [by division]"` `=2int(t+1-(2t)/(t^(2)+1)-(1)/(t^(2)+1))dt` `=2inttdt+2intdt-2int(2t)/(t^(2)+1)dt-2int(1)/(t^(2)+1)dt` `=t^(2)+2t-2log|t^(2)+1|-2tan^(-1)t+C` `=(x+1)+2sqrt(x+1)-2log|x+2|-2tan^(-1)sqrt(x+1)+C`. |
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