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

Question

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

5069 views around the world

You can reuse this answer
Creative Commons License

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.!