1.

int(3x+4)/sqrt(2x-3)dx

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`


Discussion

No Comment Found

Related InterviewSolutions