Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 25

intx/sqrt(x^2+x+1)dx
Let's rewrite the integrand by completing the square of the denominator,
=intx/sqrt((x+1/2)^2+3/4)dx
Now let's apply the integral substitution,
Let u=x+1/2
x=u-1/2
du=1dx
=int(u-1/2)/sqrt(u^2+3/4)du
=int(2u-1)/sqrt(4u^2+3)du
Now apply the sum rule,
=int(2u)/sqrt(4u^2+3)du-int1/sqrt(4u^2+3)du
=2intu/sqrt(4u^2+3)du-int1/sqrt(4u^2+3)du
Now let's evaluate the first integral by applying the integral substitution,
Let v=4u^2+3
dv=8udu
intu/sqrt(4u^2+3)du=int1/(8sqrt(v))dv
=1/8intv^(-1/2)dv
=1/8(v^(-1/2+1))/(-1/2+1)
=1/8v^(1/2)/(1/2)
=2/8v^(1/2)
=1/4sqrt(v)
substitute back v=4u^2+3
=1/4sqrt(4u^2+3)
Now let's evaluate the second integral int1/sqrt(4u^2+3)du using integral substitution,
For sqrt(bx^2+a) substitute x=sqrt(a)/sqrt(b)tan(v) ,
Let u=sqrt(3)/2tan(v)
du=sqrt(3)/2sec^2(v)dv
int1/sqrt(4v^2+3)du=int(sqrt(3)/2sec^2(v))/sqrt(4(sqrt(3)/2tan(v))^2+3)dv
=int(sqrt(3)sec^2(v))/(2sqrt(3tan^2(v)+3))dv
=sqrt(3)/2int(sec^2(v))/sqrt(3tan^2(v)+3)dv
=sqrt(3)/2int(sec^2(v))/(sqrt(3)sqrt(tan^2+1))dv
=1/2int(sec^2(v))/sqrt(tan^2(v)+1)dv
Now use the identity:1+tan^2(x)=sec^2(x)
=1/2int(sec^2(v))/sqrt(sec^2(v))dv
assuming sec(v)>=0
=1/2intsec(v)dv
Now using the common integral,
intsec(v)dx=ln((sec(v)+tan(v))
=1/2(ln(sec(v)+tan(v))
Substitute back v=arctan((2u)/sqrt(3))
=1/2[ln{sec(arctan((2u)/sqrt(3)))+tan(arctan((2u)/sqrt(3))}]
=1/2[ln{sqrt(1+(4u^2)/3)+(2u)/sqrt(3)}]
int(2u-1)/sqrt(4u^2+3)du=2(1/4sqrt(4u^2+3))-1/2ln(sqrt(1+4u^2/3)+(2u)/sqrt(3))
=1/2sqrt(4u^2+3)-1/2ln(sqrt(1+(4u^2)/3)+(2u)/sqrt(3))
Substitute back u=x+1/2
=1/2sqrt(4(x+1/2)^2+3)-1/2ln(sqrt(1+(4(x+1/2)^2)/3)+(2(x+1/2))/sqrt(3))
=1/2sqrt(4(x^2+1/4+x)+3)-1/2ln(sqrt(1+4/3(x^2+1/4+x))+(2/sqrt(3))(2x+1)/2)
=1/2sqrt(4x^2+1+4x+3)-1/2ln(sqrt((3+4x^2+1+4x)/3)+(2x+1)/sqrt(3))
=1/2sqrt(4x^2+4x+4)-1/2ln(sqrt((4x^2+4x+4)/3)+(2x+1)/sqrt(3))
=1/2sqrt(4(x^2+x+1))-1/2ln((2/sqrt(3))sqrt(x^2+x+1)+(2x+1)/sqrt(3))
=sqrt(x^2+x+1)-1/2ln((2sqrt(x^2+x+1)+2x+1)/sqrt(3))
add a constant C to the solution,
=sqrt(x^2+x+1)-1/2ln((2sqrt(x^2+x+1)+2x+1)/sqrt(3))+C