How do you integrate #int sqrt(x^2+4x+5)# using trig substitutions?

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How do you integrate #int sqrt(x^2+4x+5)dx# using trig substitutions?

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Answer 1 · 2016-09-05T14:38:05

#1/2[(x+2)sqrt(x^2+4x+5)+ln|x+2+sqrt(x^2+4x+5)|]+C#.

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.!