Evaluate `int (dx)/(x{6(logx)^(2)+7logx+2}).`

1.	Evaluate `int (dx)/(x{6(logx)^(2)+7logx+2}).`
Answer» Putting `log x=t and (1)/(x) dx=dt,` we get `I=int (dx)/(x{6(logx)^(2)+7logx+2})=int(dt)/((6t^(2)+7t+2))=int(dt)/((2t+1)(3t+2)).` `Let (1)/((2t+1)(3t+2))=(A)/((2t+1))+(B)/((3t+2)).` Then,`1-=A(3t+2)+B(2t+1).` Putting `t=-(1)/(2)`in (i) , we get `A=2` Putting `t=(-2)/(3)`in (i) , we get `B=-3.` `therefore (1)/((2t+1)(3t+2))=(2)/((2t+1))-(3)/((3t+2))` `implies I=int(dt)/((2t+1)(3t+2))` `=int (dt)/((2t+1)(3t+2))` `=log \|2t+1\|-log\|3t+2\|+C` `=log\|(2t+1)/(3t+2)\|+C` `=log \|(2logx+1)/(3logx+2)\|+C`

Discussion

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