Integral $ \int \frac{1}{\sqrt[3]{(x-1)^7 (x + 1)^2}} \mathrm{d} x $

Question

I'm having trouble with the following integral: $$ \int \frac{1}{\sqrt[3]{(x-1)^7 (x + 1)^2}} \mathrm{d} x $$ I rewrite it in the form $$ \int \frac{1}{(x - 1)\sqrt[3]{(x - 1)^2(x^2 - 1)^2}}\mathrm{d} x $$ and tried substituting $ x = \cosh \theta$. But, it doesn't seem to simplify to anything nice. I also tried $x = \cosh(2 \theta)$ in the first one and $x = \sec\theta$, but they don't seem to work, either.

Answer 1

Substitute $x=\frac{1-t}{1+t}$

\begin{align} &\int \frac{1}{\sqrt[3]{(x-1)^7 (x + 1)^2}} {d} x \\ =& -\frac14\int (t^{-7/3}+ t^{-4/3})dt = \frac34 t^{-1/3}+\frac3{16}t^{-4/3}+C \end{align}

Answer 2

$$ I=\int \frac{1}{\sqrt[3]{(x-1)^7 (x + 1)^2}} \mathrm{d} x \implies \int \frac{1}{{(x-1)^{\frac{7}{3}} (x + 1)^{\frac{2}{3}}}} \mathrm{d} x $$

Here's your generalized result specific to such kinds of integrals; $$I=\int \frac{1}{(x+a)^{\frac{m}{n}}(x+b)^{\frac{p}{n}}}\,dx$$

Where $\frac{m}{n}+\frac{p}{n}=3$

Factor out $(x+a)^3$ term,

$$I=\int \frac{1}{(x+a)^{3}\left(\frac{x+b}{x+a}\right)^{\frac{p}{n}}}\,dx$$

Just to visualize;

$$I=\int \frac{1}{(x+a)^{\frac{m}{n}+\frac{p}{n}}\left(\frac{x+b}{x+a}\right)^{\frac{p}{n}}}\,dx$$

Put $\boxed{\left(\frac{x+b}{x+a}\right)=t}\implies x=\frac{b-at}{t-1}\implies x+a = \frac{a-b}{1-t}$ , $\frac{1}{a-b}dt=\frac{1}{(x+a)^2}dx\underset{\times \frac{1}{x+a}}\implies \boxed{\frac{1-t}{(a-b)^2}dt=\frac{1}{(x+a)^3}dx}$

$$I(a;b;m;p;n) = \frac{1}{(a-b)^2}\int \frac{(1-t)}{t^{\frac{p}{n}}} dt=\frac{1}{(a-b)^2}\left[\int t^{-\frac{p}{n}}\,dt-\int t^{1-\frac{p}{n}}\,dt\right]$$

$$I(a;b;m;p;n)=\frac{1}{(a-b)^2}\left[\int t^{-\frac{p}{n}}\,dt-\int t^{1-\frac{p}{n}}\,dt\right]=\frac{1}{(a-b)^2}\left[\frac{n}{n-p}t^{1-\frac{p}{n}}-\frac{n}{2n-p}t^{2-\frac{p}{n}}\right]+C$$

Back substitute for $t$,

$$I(a;b;m;p;n)=\frac{1}{(a-b)^2}\left[\frac{n}{n-p}\left(\frac{x+b}{x+a}\right)^{1-\frac{p}{n}}-\frac{n}{2n-p}\left(\frac{x+b}{x+a}\right)^{2-\frac{p}{n}}\right]+C$$

For your integral,

$$\boxed{I(-1;1;7;2;3)=\frac{3}{4}\left(\frac{x+1}{x-1}\right)^{\frac{1}{3}}-\frac{3}{16}\left(\frac{x+1}{x-1}\right)^{\frac{4}{3}}+C}$$

Perhaps it would be great to have a generalization for the below case;

$$I=\int \frac{1}{(x+a)^{\frac{m}{n}}(x+b)^{\frac{p}{n}}}\,dx$$

Where $\frac{m}{n}+\frac{p}{n}=k$

Here are two generalizations,

For $k=2$ $$I(a;b;m;p;n)=\frac{n}{(n-p)(a-b)} \left[\frac{x+b}{x+a}\right]^{\frac{n-p}{n}}+C$$

For $k=3$ $$I(a;b;m;p;n)=\frac{1}{(a-b)^2}\left[\frac{n}{n-p}\left(\frac{x+b}{x+a}\right)^{1-\frac{p}{n}}-\frac{n}{2n-p}\left(\frac{x+b}{x+a}\right)^{2-\frac{p}{n}}\right]+C$$

Observe the pattern and generalize for any $k$

Answer 3

Note that the sum of the powers of $x+1$ and $x-1$ is $-3\in\mathbb Z^-$. In such cases, a general strategy is to take any one factor common from the radical and substitute the resulting fraction in the radical as $t$. This is because a power of $-2$ of the factor taken out as common will be used to change $\mathrm dx$ into $\mathrm dt$, so it simplifies the integration.

$$\mathcal I=\int \frac{\mathrm dx}{\sqrt[3]{(x-1)^7 (x + 1)^2}}$$ $$=\int\frac{\mathrm dx}{(x-1)^3\left(\frac{x+1}{x-1}\right)^\frac23}$$

Substitute $\frac{x+1}{x-1}=1+\frac2{x-1}=t\implies\mathrm dt=\frac{-2\mathrm dx}{(x-1)^2}$:

$$\mathcal I=-\frac14\int t^{-\frac23}(t-1)\mathrm dt$$ $$=\frac14\int t^{-\frac23}-t^\frac13\mathrm dt$$ $$=\frac34t^\frac13-\frac3{16}t^\frac43+C$$ $$=\frac34\left(\frac{x+1}{x-1}\right)^\frac13-\frac3{16}\left(\frac{x+1}{x-1}\right)^\frac43+C$$

Answer 4

We want to evaluate $$I_{p,q}=\int(x-a)^{-p}(x-b)^{-q}dx. $$ Let $t=\frac{x-b}{x-a}$. Then the integral becomes $$I_{p,q}=-\int\frac{t^{-q}(t-1)^k}{(a-b)^{k+1}}dt$$ where $k=p+q-2\in\Bbb N.$ It is clear that this integral can be evaluated "easily" term by term after the binomial expansion.

Example: $p=\frac73,q=\frac23, a=1, b=-1.$ Then $k=2$ and $t=\frac{x+1}{x-1}.$ $$I_{\tfrac73,\tfrac23}(1,-1)=-\int\frac{t^{-\tfrac23}(t-1)}{2^2}dt\\=-\frac14\int(t^{-\tfrac23}-t^{\tfrac13})dt\\=\frac34 \left(\frac{x+1}{x-1}\right)^{\tfrac13}-\frac3{16}\left(\frac{x+1}{x-1}\right)^{\tfrac43}+C$$