Let #I=intsqrt(x^2+4x+5)dx#.
#sqrt(x^2+4x+5)=sqrt{(x+2)^2+1}#,so, we take the Trigo. Substn.
#x+2=tany", so that, (1) : "dx=sec^2 ydy#, &
#(2) : sqrt(x^2+4x+5)=sqrt{tan^2 y+1}=sec y#.
#:. I=intsec ysec^2 ydy=intsec^3 ydy=intu*vdy......
[u=sec y, v=sec^2 y]#
#=uintvdy-int((du)/dx*intvdy)dy..........................[IBP]#
#=sec yintsec^2 ydy-int{sec ytan yintsec^2 y dy}dy#
#=sec ytan y-int{sec ytan ytan y}dy#
#=sec ytan y-intsec ytan^2 ydy#
#=sec ytan y-intsec y(sec^2 y-1)dy#
#=secytany-intsec^3ydy+intsecydy#, i.e.,
#I=secytany-I+ln|secy+tany|#
#:. I+I=2I=secytany+ln|secy+tany|#
#:. I=1/2[secytany+ln|secy+tany|]#
#=1/2[(x+2)sqrt(x^2+4x+5)+ln|x+2+sqrt(x^2+4x+5)|]+C#.
Enjoy Maths.!
