Evaluate the integral: I = integral cube root
Evaluate the integral: $$ I = \int \sqrt[3]{\frac{(x+1)^2}{(x-1)^5}}\,dx = \int (x+1)^{2/3}(x-1)^{-5/3}\,dx. $$
Explanation of Method
The exponents $\tfrac23$ and $-\tfrac53$ sum to the integer $-1$, so the Möbius substitution $t=\dfrac{x+1}{x-1}$ rationalizes the integral: it reduces first to $\int t^{2/3}/(1-t)\,dt$, and then $v=t^{1/3}$ makes the whole thing a rational function of $v$.
- Substitute $t=(x+1)/(x-1)$.
- Substitute $v=t^{1/3}$.
- Integrate the resulting rational function.
- Back-substitute twice.
Step 1 · Substitute $t=(x+1)/(x-1)$
Solving for $x$: $x=\dfrac{t+1}{t-1}$, so $dx=\dfrac{-2}{(t-1)^2}dt$, $x+1=\dfrac{2t}{t-1}$, $x-1=\dfrac2{t-1}$. Substituting:
$$ (x+1)^{2/3}(x-1)^{-5/3}\,dx = \left(\frac{2t}{t-1}\right)^{2/3}\left(\frac2{t-1}\right)^{-5/3}\cdot\frac{-2}{(t-1)^2}\,dt = -\frac{t^{2/3}}{t-1}\,dt. $$
$$ I = \int\frac{t^{2/3}}{1-t}\,dt. $$
Step 2 · Substitute $v=t^{1/3}$
With $t=v^3$, $dt=3v^2\,dv$:
$$ I = \int\frac{v^2}{1-v^3}\cdot3v^2\,dv = 3\int\frac{v^4}{1-v^3}\,dv. $$
Step 3 · Integrate the rational function
Polynomial division gives $\dfrac{v^4}{1-v^3}=-v+\dfrac v{1-v^3}$, and partial fractions on $\dfrac v{1-v^3}=\dfrac{v}{(1-v)(v^2+v+1)}$ complete the integration:
$$ 3\int\frac{v^4}{1-v^3}\,dv = -\frac{3v^2}2 - \ln|v-1| + \frac12\ln\left(v^2+v+1\right) - \sqrt3\arctan\left(\frac{2v+1}{\sqrt3}\right) + C. $$
Step 4 · Back-substitute twice
With $v=t^{1/3}=\left(\dfrac{x+1}{x-1}\right)^{1/3}$:
Summary Table
| Substitution | Effect |
|---|---|
| $t=(x+1)/(x-1)$ | reduces to $\int t^{2/3}/(1-t)\,dt$ |
| $v=t^{1/3}$ | reduces to $3\int v^4/(1-v^3)\,dv$, a rational integral |
Additional Notes
- Major rewrite: the source's Step 5 abandoned the correct rational-substitution approach midway ("this approach is complex... let's use another substitution") and ultimately asserted a boxed answer, $-\tfrac32\sqrt[3]{(x+1)^2/(x-1)^2}+\ln|x-1|$, that fails a direct differentiation check (derivative $\approx0.897$ vs. true integrand $\approx0.794$ at $x=3$). The boxed result below completes the rational-substitution approach the source started and was verified independently.
- The domain excludes $x=1$; for $x>1$, $t=(x+1)/(x-1)>1$ and $v=t^{1/3}>1$.
- The result was verified by differentiation at $x=3,4,5$: the derivative matches the integrand $(x+1)^{2/3}(x-1)^{-5/3}$ at all three points (e.g. $\approx0.7937$ at $x=3$).
Result
The solution to the integral is, with $v=\sqrt[3]{\dfrac{x+1}{x-1}}$: