A quicker approach to the integral $\int{dx\over{(x^3+1)^3}}$ Source: A question bank on challenging integral problems for high school students.
Problem:
Evaluate the indefinite integral
$$\int{dx\over{(x^3+1)^3}}$$
Seems pretty compact but upon closer look, no suitable substitution comes to mind. I can use partial fractions but it would be VERY time consuming and altogether boring. I am unable to find an alternate solution other than partials. Can anyone lead me towards a quicker approach, because the exam I'm preparing for is time bound and I can't afford to spend much time on a single problem. Thanks!
Edit 1:
I looked upon reduction formulas. I guess we can generalise this by using:
$$I_n = \int {dx\over{(x^3+1)^n}}$$ 
$$I_n = \int{(x^3+1)^{-n}dx}$$
Will try and solve it. Maybe we can reduce it to a simpler integral!
Edit 2:
Ok so I got the reduction formula. Can anybody verify if its right, like if you've solved it?
$$I_{n+1} = \frac {x}{3n(x^3+1)^n} + \frac{(3n-1)}{3n}I_n$$
Edit 3: Now I have reached the solution nearly
take $n=2$
$$I_3 = \frac{x}{6(x^3+1)^2}+\frac{5}{6}I_2$$
now take $n=1$
$$I_2 = \frac{x}{3(x^3+1)}+\frac{2}{3}I_1 $$
now we see that $I_1$ is nothing but 
$$I_1= \int{\frac{dx}{x^3+1}}$$ 
which simplifies to 
$$I_1 = \int{\frac{dx}{(x+1)(x^2+1-x)}}$$
Now this integral is cake, solve using Partial Fractions
$$\frac{1}{(x+1)(x^2+1-x)} \equiv \frac{A}{x+1} + \frac{Bx+C}{(x^2+1-x)}$$
And eventually $I_1$ is:
http://www.wolframalpha.com/input/?i=integrate+1%2F((x%2B1)(x%5E2%2B1-x))
This way we get $I_1$. Putting in equation above we get $I_2$
Substitute $I_2$ in the equation above and obtain an expression for $I_3$ !
I'll verify ASAP :)
Final edit: Yeas! Reached the answer. Matches term to term!
Final answer is :
http://www.wolframalpha.com/input/?i=integrate+1%2F((x%2B1)(x%5E2%2B1-x))
Use the above approach or any alternatives that are more quick are welcome!
 A: We have:
$$\int\frac{dx}{(x^3 + 1)^3} = \int \frac1{x^2}\frac{x^2}{(x^3 + 1)^3}dx$$
with 
$$u = \frac1{x^2}, dv = \frac{x^2}{(x^3 + 1)^3}dx$$
this becomes:
$$\text{something } - \frac13\int\frac{1}{x^3(x^3 + 1)^2}dx$$
then:
$$\int\frac{1}{x^3(x^3 + 1)^2}dx = \int\frac{x^3 + 1 - x^3}{x^3(x^3 + 1)^2}dx = \int\frac{dx}{x^3(x^3 + 1)} - \int\frac{dx}{(x^3 + 1)^2}$$
do same trick for $\int \frac{dx}{(x^3 + 1)^2}$ to get:
$$\int\frac{1}{x^3(x^3 + 1)^2}dx = \text{something}' + \frac53 \int\frac{dx}{x^3(x^3 + 1)}$$
so this reduces to finding
$$\int\frac{dx}{x^3 + 1} $$
this is:
$$\int\frac{1 + x^3 - x^3}{x^3 + 1} dx = x - \int x \frac{x^2}{x^3 + 1}dx$$
integrate by parts and voila!
A: In order to show the reduction formula, we use differentiation.
$$
\begin{aligned}
\frac{d}{d x}\left(\frac{x}{3 n\left(x^{3}+1\right)^{n}}\right) &=\frac{\left(x^{3}+1\right)^{n}-x n\left(x^{3}+1\right)^{n-1}\left(3 x^{2}\right)}{3 n\left(x^{3}+1\right)^{2 n}} \\
&=\frac{\left(x^{3}+1\right)^{n}-3 nx^{3}\left(x^{3}+1\right)^{n-1}}{3 n\left(x^{3}+1\right)^{2 n}} \\
&=\frac{x^{3}+1-3 n x^{3}}{3 n\left(x^{3}+1\right)^{n+1}} \\
&=\frac{1}{3 n\left(x^{3}+1\right)^{n}}-\frac{x^{3}+1-1}{\left(x^{3}+1\right)^{n+1}} \\
&=\frac{1}{3 n\left(x^{3}+1\right)^{n}} - \frac{1}{\left(x^{3}+1\right)^{n}}+\frac{1}{\left(x^{3}+1\right)^{n-1}}\\&= \frac{1-3 n}{3 n\left(x^{3}+1\right)^{n}}+\frac{1}{\left(x^{3}+1\right)^{n+1}}
\end{aligned}
$$
Integrating both sides yields $$
\begin{aligned}
\frac{x}{3 n\left(x^{3}+1\right)^{n}}&=\int \frac{1-3 n}{3 n\left(x^{3}+1\right)^{n}} d x+\int \frac{1}{\left(x^{3}+1\right)^{n+1}} d x \\
I_{n+1}&=\frac{x}{3 n\left(x^{3}+1\right)^{n}}+\frac{3 n-1}{3 n} I_{n}
\end{aligned}
$$
Wish it helps.
