`int(x+1)/(sqrt(2x^(2)+x-3))dx`

Explanation

`"Let I "- int(2x+1)/(sqrt(2x^(2) +x-3))dx`
`"Let "" "2x+1=A. (d)/(dx) (2x^(2) +x-3)+B`
` =A (4x+1) +B`
comparing the coefficients of x and constant terms
`2=4A " and " 1= A+B`
`rArr " "A=1//2 " and " B=1//2`
`"Now " I=1/2int(4x+1)/(sqrt(2x^(2)+x -3)) dx +1/3 int (1)/(sqrt(2x^(2) +x-3)) dx`
`=I_(1) +I_(2) (say)`
`I_(1)=1/2 int(4x+1)/(sqrt(2x^(2)+x-3)) dx Let (2x^(2) +x-3)=t`
` =1/2 int (1)/(sqrt(t)).dt =sqrt(t) " "rArr" "4x+ 1=(dt)/(dx)`
`rArr (4x+1) dx=dt`
`=sqrt(2x^(2) +x-3)`
`"and " I_(2) =1/2 int (1)/(sqrt(2x^(2) +x-3))`
`=(1)/(2sqrt(2))int(1)/(sqrt(x^(2) +1/2x-3/2)) dx `
`=(1)/(2sqrt(2)) int(1)/(sqrt((x^(2)+(1)/(2)x+(1)/(16) )-((3)/(2)+(1)/(16))))dx`
`=(1)/(2sqrt(2))int(1)/(sqrt((x+1/4)^(2)-((5)/(4))^(2))) dx`
`=(1)/(2sqrt(2))log |x+(1)/(4)+ sqrt((x+(1)/(4))^(2)-((5)/(4))^(2))|`
` =(1)/(2sqrt(2)) log |x+(1)/(4)+ sqrt(x^(2) +(1)/(2) x-(3)/(2))|`
`=(1)/(2sqrt(2)) log |(4x+1+sqrt(16x^(2)+8x-24))/(4)|`
`=(1)/(2sqrt(2))log |4x+1+sqrt(16x^(2)+8x-24)|`
` -(1)/(2sqrt(2))log 4`
`:. I sqrt(2x^(2) +x-3) +(1)/(2sqrt(2)) log `
` |4x+1+sqrt(16x^(2) +8x-24|)+c.`
`:. -(1)/(2sqrt(2)) log 4 " is constant "`