`int sqrt(x^2 + 2x) dx` Evaluate the integral

`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`