Vtyjkyuk

Evaluate
$$
int fracmathrmdxxsqrtx^2+x+1 cdotp
$$

My attempt:
$$
I
= int fracmathrmdxxsqrtx^2+x+1
= int fracmathrmdxxsqrtleft(x + frac12right)^2 + left( fracsqrt32 right)^2
$$

I thought completing the square would bring the integrand into some form, but it did not. Please help.

Hint: I know a way. The integral of form $$intfracdx(x-n)^msqrtax^2+bx+c$$ could be solved by taking $x-n=1/t.$. If you do this, then the whole integral will change to a integral of form $intfracP(x)dxsqrtax^2+bx+c$. Try this. Tell me if you can do the last one or not.

First use the substitution $x=frac1t$ to get
$$int fracmathrmdxxsqrtx^2+x+1 =-int fracmathrmdtsqrt1+t+t^2.$$
Now complete the squares and use a tan substitution.

Try $x=1/t$. The integrand reduces to a known form.

For integrals of the form $$dfrac1(x+a)sqrt(x+b)^2+c^2,$$

Choose $x+b= c
tan y,$ to reach at an integral of the form $$dfrac1Acos y+Bsin y$$

Now $Acos y+Bsin y=sqrtA^2+B^2sin (y+arctan dfrac AB)=sqrtA^2+B^2cos(y-arctandfrac AB)$

$$I=classsteps-nodecssIdsteps-node-12displaystyleintdfrac1xsqrtleft(2x+1right)^2+3,mathrmdx$$

Substitute $u=2x+1$

$$I=2displaystyleintdfrac1left(u-1right)sqrtu^2+3,mathrmdu$$

Substitute $u=sqrt3tanleft(vright)$
$$I=2displaystyleintdfracsqrt3sec^2left(vright)left(sqrt3tanleft(vright)-1right)sqrt3tan^2left(vright)+3,mathrmdv$$
$$I=2displaystyleintdfractan^2left(fracv2right)+1left(1-tan^2left(fracv2right)right)left(frac2cdotsqrt3tanleft(fracv2right)1-tan^2left(fracv2right)-1right),mathrmdv$$

Substitute $w=tanleft(dfracv2right)$

$$I=classsteps-nodecssIdsteps-node-24displaystyleintdfrac1w^2+2cdotsqrt3w-1,mathrmdw$$

$$I=4displaystyleintdfrac1left(w+sqrt3-2right)left(w+sqrt3+2right),mathrmdw$$