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int(3x+4)/sqrt(2x-3)dx |
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Answer» Solution :`INT(3x+4)/sqrt(2x-3)DX` [Put 2x-3=t^2 Then 2dx=2tdt or dx=tdt] =`int(3((t^2+3)/2)+4)/t .tdt` =`3/2intt^2dt+17/2intdt=3/2.t^3/3+17/2t+C` =`1/2t^3+17/2t+C` =`1/2(2x-3)^(2/3) +17/2(2x-3)^(1/2) +C` =`(x+7)sqrt(2x-3)dx` |
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