I have an exercise where I need to integrate $\displaystyle \int \dfrac{81x-162}{x^4 + 81x^2} dx$ exclusively by the method of integration by parts. I don't know what to do, all I can do is integrate this by performing partial fractions decompositions, which I can't use. When I try to integrate by parts, the $\int v du$ term is way more complicated than the original integral, doesn't matter what I choose for $u$ or $dv$.
How can I integrate this by the integration by parts method?
Thanks.
Best Answer
Let's rewrite the integrand as follows:
$$\frac{81x-162}{x^4+81x^2}=(81x-162)\cdot\frac{1}{x^4+81x^2}$$
Differentiating $1/(x^4+81x^2)$ will definitely make things harder than they need to be, so we'll differentiate $81x-162$ instead.
To use integration by parts this way, we need to find an antiderivative for $1/(x^4+81x^2)$, so we begin by evaluating the integral
$$\int\frac{1}{x^4+81x^2}dx$$
Integrals of rational functions can be evaluated in a systematic fashion by using a partial fraction decomposition. This integral, however, has a much better solution:
\begin{align*} \int\frac{1}{x^4+81x^2}dx &= \int\frac{1}{x^4\left(1+\frac{81}{x^2}\right)}dx\\ &= -\int-\frac{1}{x^2}\cdot\left(\frac{1}{x}\right)^2\cdot\frac{1}{1+81\left(\frac{1}{x}\right)^2}dx\\ \end{align*}
Making the substitution $u=\frac{1}{x}$ and noting that $du=-\frac{1}{x^2}dx$ gives
\begin{align*} -\int-\frac{1}{x^2}\cdot\left(\frac{1}{x}\right)^2\cdot\frac{1}{1+81\left(\frac{1}{x}\right)^2}dx &= -\int\frac{u^2}{1+81u^2}du\\ &= -\frac{1}{81}\int\frac{81u^2}{1+81u^2}du\\ &= -\frac{1}{81}\int\frac{1+81u^2-1}{1+81u^2}du\\ &= -\frac{1}{81}\int\left(\frac{1+81u^2}{1+81u^2}-\frac{1}{1+81u^2}\right)du\\ &= -\frac{1}{81}\int\left(1-\frac{1}{1+9^2u^2}\right)du\\ &= \frac{1}{81}\int\frac{1}{9^2\left(\frac{1}{9^2}+u^2\right)}du-\frac{1}{81}\int 1\text{ }du\\ &= \frac{1}{81\cdot 9^2}\cdot\frac{1}{1/9}\tan^{-1}\left(\frac{u}{1/9}\right)-\frac{u}{81}+C\\ &= \frac{9}{81\cdot 9^2}\tan^{-1}(9u)-\frac{u}{81}+C\\ &= \frac{1}{729}\tan^{-1}\left(\frac{9}{x}\right)-\frac{1}{81x}+C\\ &= \frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}+C \end{align*}
We can now (finally) evaluate the original integral by parts. Since $\frac{d}{dx}(81x-162)=81$, we have
\begin{align*} \int\frac{81x-162}{x^4+81x^2}dx &= (81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\int 81\cdot\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]dx\\ &= (81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\int\left[\frac{1}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{x}\right]dx\\ &= (81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\int\frac{1}{9}\cot^{-1}\left(\frac{x}{9}\right)dx+\int\frac{1}{x}dx\\ &= (81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\int\cot^{-1}(u)du+\ln|x|\\ &= (81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\left[u\cot^{-1}(u)+\frac{1}{2}\ln\left(1+u^2\right)\right]+\ln|x|+C\\ &= (81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\frac{x}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{2}\ln\left(1+\frac{x^2}{81}\right)+\ln|x|+C\\ \end{align*}
Note: in the last part of the solution, we made the substitution $u=x/9$ and used the standard integral
$$\int\cot^{-1}(u)\text{ }du=u\cot^{-1}(u)+\frac{1}{2}\ln(1+u^2)$$
Hope this helps!
Edit: as was pointed out by a comment on my answer, the resulting expression simplifies quite nicely:
$$(81x-162)\left[\frac{1}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{81x}\right]-\frac{x}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{2}\ln\left(1+\frac{x^2}{81}\right)+\ln|x|+C\\$$ \begin{align*} &= \frac{81x-162}{729}\cot^{-1}\left(\frac{x}{9}\right)-\frac{81x-162}{81x}-\frac{x}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{2}\ln\left(1+\frac{x^2}{81}\right)+\ln|x|+C\\ &= \frac{x}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{2}{9}\cot^{-1}\left(\frac{x}{9}\right)-1+\frac{2}{x}-\frac{x}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{2}\ln\left(1+\frac{x^2}{81}\right)+\ln|x|+C\\ &= -\frac{2}{9}\cot^{-1}\left(\frac{x}{9}\right)-1+\frac{2}{x}-\frac{1}{2}\ln\left(1+\frac{x^2}{81}\right)+\ln|x|+C\\ &= \frac{2}{x}-\frac{2}{9}\cot^{-1}\left(\frac{x}{9}\right)-\frac{1}{2}\ln\left(1+\frac{x^2}{81}\right)+\ln|x|+C \end{align*}
The $-1$ was absorbed into the $C$ in the last step.