Integration of $\int (\frac{1-x}{1+x})^{\frac{1}{3}}$

Question

I've been trying to integrate this for a long time but can't.

$$\int \left(\frac{1-x}{1+x}\right)^{\frac{1}{3}}$$

I tried assuming $\frac{1-x}{1+x}=t^3$ , also tried integration by parts but it gets stuck in a loop.

Answer 1

Hint. The obvious substitution here is $x=\frac{1-u}{1+u}$. Using this, we find:

$$\begin{align} \int\left(\frac{1-x}{1+x}\right)^{\frac13}\,\mathrm{d}x &=\int u^{\frac13}\cdot\frac{(-2)\,\mathrm{d}u}{(1+u)^2}\\ &=-2\int\frac{u^{\frac13}}{(1+u)^2}\,\mathrm{d}u. \end{align}$$

Next, substitute $u=t^3$:

$$\begin{align} -2\int\frac{u^{\frac13}}{(1+u)^2}\,\mathrm{d}u &=-6\int\frac{t^3}{(1+t^3)^2}\,\mathrm{d}t.\\ \end{align}$$

This transforms the integrand to a rational function, so the rest of the problem is now just a matter of computation.

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EDIT:

Since the integration of the rational function arrived at in the hint above isn't necessarily as straightforward as I initially thought, I'll go ahead and complete the solution.

First note that $$\frac{d}{dt}\left[\frac{2}{1+t^3}\right]=\frac{-6\,t^2}{(1+t^3)^2}.$$

Thus, integrating by parts we have:

$$\begin{align} \int\frac{-6\,t^3}{(1+t^3)^2}\,\mathrm{d}t &=\int t\left[\frac{-6\,t^2}{(1+t^3)^2}\right]\,\mathrm{d}t\\ &=\frac{2\,t}{1+t^3}-2\int\frac{\mathrm{d}t}{1+t^3}. \end{align}$$

The integral $\int\frac{\mathrm{d}t}{1+t^3}$ can be done by partial fractions.

$$\frac{1}{1+t^3}=\frac{1}{3(1+t)}+\frac{2-t}{3(1-t+t^2)}.$$

For convenience, we may rewrite the second term as:

$$\begin{align} \frac{2-t}{3(1-t+t^2)} &=-\frac{-2+t}{3(1-t+t^2)}\\ &=-\frac{-4+2t}{6(1-t+t^2)}\\ &=-\frac{(-1+2t)-3}{6(1-t+t^2)}\\ &=-\frac{-1+2t}{6(1-t+t^2)}+\frac{3}{6(1-t+t^2)}\\ &=-\frac{-1+2t}{6(1-t+t^2)}+\frac{1}{2(1-t+t^2)}. \end{align}$$

Thus,

$$\begin{align} \int\frac{\mathrm{d}t}{1+t^3} &=\frac13\int\frac{\mathrm{d}t}{1+t}-\frac16\int\frac{-1+2t}{1-t+t^2}\,\mathrm{d}t+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\ &=\frac13\ln{(1+t)}-\frac16\ln{(1-t+t^2)}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\ &=\frac26\ln{(1+t)}-\frac16\ln{(1-t+t^2)}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\ &=\frac16\ln{\left[(1+t)^2\right]}-\frac16\ln{(1-t+t^2)}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\ &=\frac16\ln{\left[\frac{(1+t)^2}{1-t+t^2}\right]}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}\\ &=\frac16\ln{\left[\frac{(1+t)^3}{1+t^3}\right]}+\frac12\int\frac{\mathrm{d}t}{1-t+t^2}. \end{align}$$

Finally, the integral $\frac12\int\frac{\mathrm{d}t}{1-t+t^2}$ can be found most easily by completing the square in the denominator. We have,

$$\begin{align} t^2-t+1 &=\left(t^2-t+\frac14\right)+\left(1-\frac14\right)\\ &=\left(t-\frac12\right)^2+\frac34\\ &=\frac14\left(2t-1\right)^2+\frac34\\ &=\frac34\left(\frac{2t-1}{\sqrt{3}}\right)^2+\frac34\\ &=\frac34\left[\left(\frac{2t-1}{\sqrt{3}}\right)^2+1\right]. \end{align}$$

Thus, a convenient substitution would be $\frac{2t-1}{\sqrt{3}}=r\implies t=\frac{\sqrt{3}\,r+1}{2}\implies \mathrm{d}t=\frac{\sqrt{3}}{2}\,\mathrm{d}r$:

$$\begin{align} \frac12\int\frac{\mathrm{d}t}{1-t+t^2} &=\frac23\int\frac{\mathrm{d}t}{\left(\frac{2t-1}{\sqrt{3}}\right)^2+1}\\ &=\frac23\int\frac{\frac{\sqrt{3}}{2}\,\mathrm{d}r}{r^2+1}\\ &=\frac{1}{\sqrt{3}}\int\frac{\mathrm{d}r}{r^2+1}\\ &=\frac{1}{\sqrt{3}}\arctan{\left(r\right)}+\color{grey}{constant}\\ &=\frac{1}{\sqrt{3}}\arctan{\left(\frac{2t-1}{\sqrt{3}}\right)}+\color{grey}{constant}. \end{align}$$

Hence,

$$\int\frac{\mathrm{d}t}{1+t^3}=\frac16\ln{\left[\frac{(1+t)^3}{1+t^3}\right]}+\frac{1}{\sqrt{3}}\arctan{\left(\frac{2t-1}{\sqrt{3}}\right)}+\color{grey}{constant}.$$

Hence,

$$\int\frac{-6\,t^3}{(1+t^3)^2}\,\mathrm{d}t=\frac{2\,t}{1+t^3}-\frac13\ln{\left[\frac{(1+t)^3}{1+t^3}\right]}-\frac{2}{\sqrt{3}}\arctan{\left(\frac{2t-1}{\sqrt{3}}\right)}+\color{grey}{constant}.$$

Finally, to obtain the integral $\int\left(\frac{1-x}{1+x}\right)^{\frac13}\,\mathrm{d}x$, we simply substitute back $t=\sqrt[3]{u}=\sqrt[3]{\frac{1-x}{1+x}}$ into the expression for the integral $\int\frac{-6\,t^3}{(1+t^3)^2}\,\mathrm{d}t$ stated above. This comes out to,

$$\int\left(\frac{1-x}{1+x}\right)^{\frac13}\,\mathrm{d}x=(1-x)^{1/3}(1+x)^{2/3}-\frac13\ln{\left[\frac{(1+x)\left(1+\sqrt[3]{\frac{1-x}{1+x}}\right)^3}{2}\right]}-\frac{2}{\sqrt{3}}\arctan{\left(\frac{2\sqrt[3]{\frac{1-x}{1+x}}-1}{\sqrt{3}}\right)}+\color{grey}{constant}.$$

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Note: As an aside, it's interesting to observe that by combining the anti-derivative I found with the anti-derivative Shadock gave in terms of hypergeometric functions, we can then deduce a closed-form for the Gauss hypergeometric function $_2F_1(\frac{2}{3}, \frac{2}{3}, \frac{5}{3}, \frac{x+1}{2})$.

Answer 2

The answer of wolfram alpha is

$ \int (\frac{1-x}{1+x})^{1/3} dx = \frac{(2^{1/3}*(1-x)^{2/3}*_2F_1(\frac{2}{3}, \frac{2}{3}, \frac{5}{3}, \frac{x+1}{2})-2 x+2)}{2*(\frac{1-x}{1+x})^{2/3}}+constant $

with F the hypergeometric function... So good luck for you ^^

Yann

Answer 3

Hint: $(\frac{(1-x)}{(1+x)^{1/3}}=(\frac 1{(1+x))^{1/3}}-(\frac{(x)}{(1+x))^{1/3}} $[can you break $(\frac{(x)}{(1+x))^{1/3}} $ more to get a suitable form?]

After edit Hint:

Try with trigonometric function.