8

How to evaluate the following pseudo-elliptic integral?

$$\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx$$

I think that I should make $(x+1)^4$ and substract something under the square root. I got $(x+1)^4 -x^4-3x^3-5x^2-3x-1$ but I have no idea what I should do with summands.

metamorphy
  • 39,111
James
  • 125

4 Answers4

8

Apply the Euler substitution $\sqrt{x^3+x^2+x}=xt$. Then, $t^2=x+1+\frac1x$, $dx= \frac{2tx^2}{x^2-1}dt$ and $$\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}}dx =\int \frac2{t^2+1}dt=2\tan^{-1}t+C $$

Quanto
  • 97,352
  • I'm confused. we use Euler substitution when we have (ax^2+bx+c) under square root. arcsin((x^2+1)/(x+1)^2)+C is answer in the book – James Jun 06 '23 at 03:50
  • @James - Note that $$2\tan^{-1}\frac{\sqrt{x^3+x^2+x}}x-\sin^{-1}\frac{x^2+1}{(x+1)^2}=\frac\pi2$$ They differ by a constant – Quanto Jun 06 '23 at 04:06
  • Thanks a lot!!! – James Jun 06 '23 at 11:03
2

Take the argument from the book as the argument transformation and consider the square (as one should do always with elliptic integrals, signs may be considered later on).

In Mathematica in order to avoid trival transfer errors

       Solve[(x^2 + 1)/(x + 1)^2 == y, x] // FullSimplify
       {{x->(y+Sqrt[-1+2 y])/(1-y)},{x->(-y+Sqrt[-1+2 y])/(-1+y)}}

  ((dx (x - 1)/((x + 1) Sqrt[x] Sqrt[x^2 + x + 1])))^2 //. 
    {x -> (y + Sqrt[-1 + 2 y])/(1 - y), 
     dx -> D[(y + Sqrt[-1 + 2 y])/(1 - y), y] dy} // FullSimplify

     dy^2/(1 - y^2)

This yields the $\arcsin$ with domain and inversion of the map in the complex domain open to discussion. Real line integrals are always given as half definite integrals with lower or upper limits pinned to one of the branch points of the square roots. The indefinite integrals have an algebraical local meaning only.

Roland F
  • 2,098
2

Hint: $$\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx\\~\\＝\int\frac{(x-1)(x＋1)}{(x+1)^2\sqrt{x^3+x^2+x}}\,dx\\~\\＝\int\;\frac {x^2-1}{(x^2＋2x＋1)\sqrt{x^3+x^2+x}}dx\\~\\＝\int\;\frac {x^2-1}{(x^3＋2x^2＋x)\sqrt {x＋1＋\frac {1}{x}}}dx\\~\\\int\;\frac {1-\frac {1}{x^2}}{(x＋2＋\frac {1}{x})\sqrt {x＋1＋\frac {1}{x}}}dx$$

Now : let u=$x+\frac{1}{x}$

Mostafa
  • 1,479
  • 14
1

Substitute $x=\dfrac{1-y}{1+y}$ then $z=y^2+1$ :

$$\begin{align*} & \int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx \tag1 \\ &= \int \frac{2y}{\sqrt{\frac{(1-y)^3}{(1+y)^3} + \frac{(1-y)^2}{(1+y)^2} + \frac{1-y}{1+y}}} \, \frac{dy}{(1+y)^2} \\ &= \int \frac{2y}{\sqrt{3-2y^2-y^4}} \, dy \\ &= \int \frac{2y}{\sqrt{4-\left(y^2+1\right)^2}} \, dy \\ &= \int \frac{dz}{\sqrt{4-z^2}} \tag2 \end{align*}$$

user170231
  • 19,334