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I encountered a very difficult problem, to calculate the answer of this formula: $$ \int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x $$ Can you help me to find out how it solved?

Masquentin
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    Do tell whatever you tried while solving the problem. HINT: Substitute $x = tan(t)$. The integrand will reduce to $cot^8t$ which can be integrated using reduction formulas. – Scilife Nov 03 '20 at 07:50
  • Have you tried to use partial fractions? – Aaron Nov 03 '20 at 07:50
  • Thanks! I think I stuck into the wrong exchange method. I really appreciate your help! – Masquentin Nov 03 '20 at 13:38

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$$\frac{1}{x^8(1+x^2)}=\frac{1+x^2-x^2}{x^8(1+x^2)}=\frac{1}{x^8}-\frac{1}{x^6(1+x^2)}=\cdots$$ and so on, finally you will get: $$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$

Riemann
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Using partial fraction decomposition (left as an exercise for the reader), notice that $$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$ Therefore, $$\int \frac{1}{x^8(x^2 + 1)} ~ dx = \int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right] ~ dx.$$ From here, one can notice that the first $4$ terms have simple anti-derivatives, and the anti-derivative of the last term is $\arctan x.$ One can do a sanity check by computing the derivative of $\arctan x$ to see that it is indeed the case. Therefore, we have that $$\int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right] ~ dx = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - \frac{1}{7x^7} + \arctan x + C.$$

trivial math is difficult
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    welcome to mse! Considering the fact that the OP has literally shown no effort,i believe such questions shouldn.t be answered , or to a limit a hInt is given rather than a complete question,although your solution is nice – Albus Dumbledore Nov 03 '20 at 08:07
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We need to integrate: $$ I = \int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x $$ Continuing from my comment, substitute $x = tan\theta$. Then, $dx = sec^2\theta. d\theta$. We also know that $1+ tan^2\theta = sec^2\theta$. Substituting back into the integral, we get:

$$I =\int \frac{sec^2\theta}{tan^{8}\theta \left(1+tan^{2}\theta\right)} \ \mathrm{d}\theta$$ $$ \implies I = \int \frac{1}{tan^{8}\theta} \ \mathrm{d}\theta$$ $$ \implies I =\int \cot^8\theta \ \mathrm{d}\theta$$

We can now handle this integral quite easily by repeatedly using the formula, $cot^2\theta = cosec^2\theta - 1$.

We finally get, $$ I = -\frac{cot^7\theta}{7}+\frac{cot^5\theta}{5}-\frac{cot^3\theta}{3}+cot\theta\ + \theta + C$$

Now substitute back $\theta = tan^{-1}x$ for your final answer!

Scilife
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  • Thanks! I think I stuck into the wrong exchange method. I really appreciate your help! – Masquentin Nov 03 '20 at 13:39
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    Another interesting observation is that the integral follows a recurrence relation: for $I_n = \int \cot^n \theta d\theta$, $I_n = -\frac{\cot^n-1}{n-1} - I_{n-2}$. This means that you can find the answer easily, even with a large exponent eg $\int \cot^{200}x dx$ – Aniruddha Deb Dec 04 '20 at 02:46