$\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx$

Question

How would one go about integrating $\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx$

I tried rationalizing and then doing partial fractions but got something really ugly and IBP doesn't work too well either.

I edited your question. Is this the function you meant? – Cornman Oct 26 '17 at 02:44 — Cornman, Oct 26 '17 at 02:44
Yes sir, thank you so much! – WhatsDUI Oct 26 '17 at 02:44 — WhatsDUI, Oct 26 '17 at 02:44
In case you want to double check your answer, check out. – Quang Hoang Oct 26 '17 at 02:50 — Quang Hoang, Oct 26 '17 at 02:50

score 4 · Accepted Answer · answered Oct 26 '17 at 03:14

4

Let $x=\dfrac1u$ then \begin{align} I &= \int\dfrac{-u}{\sqrt{u^2+u+1}}du \\ &= -\int\dfrac{2u+1}{2\sqrt{u^2+u+1}}du+\dfrac12\int\dfrac{1}{\sqrt{(u+\frac12)^2+\frac34}}du \\ &= -\sqrt{u^2+u+1}+\dfrac12\operatorname{arcsinh}\dfrac{2u+1}{\sqrt{3}}+C \\ &= -\dfrac{\sqrt{x^2+x+1}}{x}+\dfrac12\operatorname{arcsinh}\dfrac{x+2}{x\sqrt{3}}+C \end{align}

answered Oct 26 '17 at 03:14

Nosrati

29,995

The only sensible answer so far. +1 – Jack Tiger Lam Oct 26 '17 at 04:04
Yes, very nice! +1 – Michael Rozenberg Oct 26 '17 at 04:58

score 3 · Answer 2 · answered Oct 26 '17 at 02:50

3

$x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}$.

Thus, the substitution $x+\frac{1}{2}=\frac{\sqrt3}{2}\tan{t}$ must help.

answered Oct 26 '17 at 02:50

Michael Rozenberg

194,933

score 1 · Answer 3 · answered Oct 26 '17 at 02:49

1

Hint:

$\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx=\int \frac{1}{x^2\sqrt{(x+\frac12)^2+\frac34}}\, dx$

Now substitute $u=x+\frac12$

answered Oct 26 '17 at 02:49

Cornman

11,065
4
30
57

$\int \frac{1}{x^2\sqrt{x^2+x+1}}\, dx$

3 Answers3