Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 27

intsqrt(x^2+2x)dx
Let's rewrite the integrand by completing the square for x^2+2x ,
=intsqrt((x+1)^2-1)dx
Now apply integral substitution,
Let u=x+1,
=>du=1dx
=intsqrt(u^2-1)du
Now again apply the integral substitution,
Let u=sec(v),
=>du=sec(v)tan(v)dv
=intsqrt(sec^2(v)-1)sec(v)tan(v)dv
Now use the identity: sec^2(x)=1+tan^2(x)
=intsqrt(1+tan^2(v)-1)sec(v)tan(v)dv
=intsqrt(tan^2(v))sec(v)tan(v)dv
assuming tan(v)>=0,sqrt(tan^2(v))=tan(v)
=inttan^2(v)sec(v)dv
Using the identity: tan^2(x)=sec^2(x)-1
=int(sec^2(v)-1)sec(v)dv
=int(sec^3(v)-sec(v))dv
Apply the sum rule,
=intsec^3(v)dv-intsec(v)dv
Now let's evaluate the first integral by applying the integral reduction,
intsec^n(x)=(sec^(n-1)(x)sin(x))/(n-1)+(n-2)/(n-1)intsec^(n-2)(x)dx
intsec^3(v)dv=(sec^2(v)sin(v))/2+(3-2)/(3-1)intsec(v)dv
intsec^3(v)dv=(sec^2(v)sin(v))/2+1/2intsec(v)dv
Now use the common integral: intsec(x)dx=ln|sec(x)+tan(x)|
intsec^3(v)dv=(sec^2(v)sin(v))/2+1/2(ln|sec(v)+tan(v)|)
Now plug back the above integral and the common integral,
=(sec^2(v)sin(v))/2+1/2(ln|sec(v)+tan(v)|)-ln|sec(v)+tan(v)|
=(sec^2(v)sin(v))/2-1/2(ln|sec(v)+tan(v)|)
=sin(v)/(2cos^2(v))-1/2(ln|sec(v)+tan(v)|)
=(sec(v)tan(v))/2-1/2(ln|sec(v)+tan(v)|)
Now substitute back: u=sec(v). u=(x+1)
=>v=arcsec(u)
=>v=arcsec(x+1)
=(sec(arcsec(x+1))tan(arcsec(x+1)))/2-1/2(ln|sec(arcsec(x+1))+tan(arcsec(x+1))|)
Now tan(arcsec(x+1))=sqrt((x+1)^2-1)
=((x+1)sqrt((x+1)^2-1))/2-1/2(ln|(x+1)+sqrt((x+1)^2-1)|) + C