Evaluate: $\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}$

Question

Evaluate $$ \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} \cdotp $$

My attempt: $$ I = \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} = \int \frac{\mathrm{d}x}{x\sqrt{\left(x + \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}} $$

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

Answer 1

Hint: I know a way. The integral of form $$\int\frac{dx}{(x-n)^m\sqrt{ax^2+bx+c}}$$ could be solved by taking $x-n=1/t.$. If you do this, then the whole integral will change to a integral of form $\int\frac{P(x)dx}{\sqrt{ax^2+bx+c}}$. Try this. Tell me if you can do the last one or not.

Answer 2

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

Answer 3

First use the substitution $x=\frac{1}{t}$ to get $$\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} =-\int \frac{\mathrm{d}t}{\sqrt{1+t+t^2}}.$$ Now complete the squares and use a tan substitution.

Answer 4

$$I=\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{x\sqrt{\left(2x+1\right)^2+3}}\,\mathrm{d}x$$

Substitute $u=2x+1$

$$I=2{\displaystyle\int}\dfrac{1}{\left(u-1\right)\sqrt{u^2+3}}\,\mathrm{d}u$$

Substitute $u=\sqrt{3}\tan\left(v\right)$ $$I=2{\displaystyle\int}\dfrac{\sqrt{3}\sec^2\left(v\right)}{\left(\sqrt{3}\tan\left(v\right)-1\right)\sqrt{3\tan^2\left(v\right)+3}}\,\mathrm{d}v$$ $$I=2{\displaystyle\int}\dfrac{\tan^2\left(\frac{v}{2}\right)+1}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)\left(\frac{2\cdot\sqrt{3}\tan\left(\frac{v}{2}\right)}{1-\tan^2\left(\frac{v}{2}\right)}-1\right)}\,\mathrm{d}v$$

Substitute $w=\tan\left(\dfrac{v}{2}\right)$

$$I=\class{steps-node}{\cssId{steps-node-2}{4}}{\displaystyle\int}\dfrac{1}{w^2+2\cdot\sqrt{3}w-1}\,\mathrm{d}w$$

$$I=4{\displaystyle\int}\dfrac{1}{\left(w+\sqrt{3}-2\right)\left(w+\sqrt{3}+2\right)}\,\mathrm{d}w$$

Answer 5

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 $$\dfrac1{A\cos y+B\sin y}$$

Now $A\cos y+B\sin y=\sqrt{A^2+B^2}\sin (y+\arctan \dfrac AB)=\sqrt{A^2+B^2}\cos(y-\arctan\dfrac AB)$

Answer 6

Per the Euler substitution $\sqrt{x^2+x+1}=(x+\frac12)t+\frac{\sqrt3}2$

$$ \int \frac{{d}x}{x\sqrt{x^2+x+1}} = \int\frac{dt}{(\frac{t+\sqrt3}2)^2-1} = -2\tanh^{-1}\frac{\sqrt{x^2+x+1}-\sqrt3x}{2x+1} $$