`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`
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