Calculus Integral from Partial Fractions

Question

When you have an irreducible quadratic factor repeated you can get integrals that look like $\int \dfrac{dx}{(x^2+a)^m}$, where $m>1$, integer, and $a>0$. What is the best way to integrate this function? Is there more than one way?

Answer 1

The recurrence relation derived by Jack D'Aurizio in his answer above can actually be solved explicitly for $I_m$. First let me quote it below, but changing the recurrence index to $n$:

$$ I_{n+1} = \frac{2n+1}{2n}\,I_{n} + f_n(t) \qquad (n \ge 1) \qquad (1) $$ where $$ \begin{align} I_0 &= t \\[0.05in] I_1 &= \arctan(t) \\[0.05in] f_n(t) &\equiv \frac{1}{2n}\frac{t}{(1+t^2)^n} \qquad (n \ge 1) \end{align} $$

Now define the quantity

$$ P_{n} \equiv \prod_{k\,=\,1}^{n} \left(\frac{2k+1}{2k}\right) = \prod_{k\,=\,1}^{n} \left(1 + \frac{1}{2k}\right) $$

Then, dividing (1) by $P_n$, we find $$ \frac{I_{n+1}}{P_{n}} - \left(\frac{2n+1}{2n}\right)\frac{I_{n}}{P_{n}} = \frac{f_n}{P_{n}} $$

Since $$ P_{n} = \left(\frac{2n+1}{2n}\right)P_{n-1} $$ we have $$ \frac{I_{n+1}}{P_{n}} - \frac{I_{n}}{P_{n-1}} = \frac{f_n}{P_{n}} $$

Now sum both sides on $n$: $$ \sum_{n\,=\,2}^{m-1} \left( \frac{I_{n+1}}{P_{n}} - \frac{I_{n}}{P_{n-1}} \right)= \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}} $$

The LHS telescopes, so $$ \frac{I_{m}}{P_{m-1}} - \frac{I_{2}}{P_{1}} = \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}} $$ and $$ I_{m} = P_{m-1} \left( \frac{I_{2}}{P_{1}} + \sum_{n\,=\,2}^{m-1} \frac{f_n}{P_{n}} \right) = P_{m-1} \left( \frac{I_{2}}{P_{1}} - \frac{f_1}{P_{1}} + \sum_{n\,=\,1}^{m-1} \frac{f_n}{P_{n}} \right) = P_{m-1} \left( I_1 + \sum_{n\,=\,1}^{m-1} \frac{f_n}{P_{n}} \right) $$ where (1) was used in the last equality to remove $I_2$.

Hence,

$$ I_{m} = P_{m-1} \left( \arctan(t) + \sum_{n\,=\,1}^{m-1} \frac{t\,(1+t^2)^{-n}}{2n\,P_{n}} \right) \qquad (m \ge 2) $$

Answer 2

Since $a > 0$, let $a = b^2$. Then let $$x = b\tan\theta$$ $$dx = b\sec^2\theta\,d\theta$$

so that

$$ \int \dfrac{dx}{(x^2+b^2)^m} = \int \dfrac{b\sec^2\theta\,d\theta}{b^{2m}\sec^{2m}\theta} = b^{1-2m}\int \cos^{2(m-1)}\theta\,d\theta $$

This can now be integrated by parts repeatedly.

Answer 3

Let: $$ I_m = \int \frac{dt}{(1+t^2)^m}.$$ We have $I_0=t, I_1 = \arctan t$ and: $$ I_{m+1}=\int \frac{(1+t^2)-t^2}{(1+t^2)^{m+1}}\,dt = I_m-\int \frac{t}{2}\cdot\frac{2t}{(1+t^2)^{m+1}}\,dt$$ hence, through integration by parts: $$ I_{m+1} = I_m+\frac{t}{2m}\cdot\frac{1}{(1+t^2)^m}+\frac{I_m}{2m} $$ and we have the recursive formula:

$$ I_{m+1} = \frac{2m+1}{2m}\,I_m + \frac{1}{2m}\cdot\frac{t}{(1+t^2)^m}. $$

Answer 4

Hint: Evaluate $I(a)=\displaystyle\int\frac{dx}{x^2+a}$ , and then differentiate both sides $m-1$ times with regard to a.