If $\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c}$, then find $a,b,c$.

If

$$\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c}$$

then find $$a, b, c$$.

Now, using partial fractions I calculated

$$a = 62-10\ln (2) \qquad\qquad b = -15 \qquad\qquad c = 20$$

but it took me more than 45 minutes to do all the work. The question was asked in an MCQ exam where only 4-5 minutes are available. I am probably missing something which can help to solve it. Thanks!

Note it was asked in an exam for students of grade 12 so I basically don't know very complex integrations thus I am searching for integration via elementary functions only.

• Looking at this, I kind of suspect it was meant as a reward for those who work quickly. I don't think you can do much better than partial fractions. – user361424 Oct 29 '16 at 6:37
• I don't understand how you could use that the right hand side is of that form... are there are least sone constrains on $a,b,c$ you forgot to mention? Otherwise that $\pi$ is totally sensless. Did you have some possible answers to choose from? – b00n heT Oct 29 '16 at 6:38
• Exactly but even the best ones will at least take 10 mins to solve i think – Archis Welankar Oct 29 '16 at 6:39
• if you want to try something really fancy (which is a also a fast method if you know what u are doing): Integrate $f(z)=\frac{\log(1-z)(z^3+3)}{z^6(z^2+1)}$ around a keyhole contour in the complex plane – tired Oct 29 '16 at 10:36
• I also want to complain against the proposed format for the answer. Unless there is some requirement on $a$, $b$ and $c$ you could always use $b=0$, $c=1$ and $a=$ whatever the integral is. Things like this make people think mathematical expressions are just abstract symbols on a page instead of actual numbers. – Javier Oct 29 '16 at 14:17

Use the change of variable $x\mapsto x^{-1}$ to turn your integral into $$\int_0^1\frac{x^3(1+3x^3)}{1+x^2}\,dx=\int_0^1\frac{x^3}{1+x^2}\,dx+3\int_0^1\frac{(1+x^2-1)^3}{1+x^2}\,dx\\ =\frac{1}{2}\int_0^1\frac{x\,dx}{1+x}+3\int_0^1(1+x^2)^2\,dx-9\int_0^1(1+x^2)\,dx+9\int_0^1\,dx-3\int_0^1\frac{\,dx}{1+x^2}\\ =\frac{1}{2}\int_1^2\frac{x-1}{x}\,dx+3\int_0^1\,dx-3\int_0^1x^2\,dx+3\int_0^1x^4\,dx-3\int_0^1\frac{\,dx}{1+x^2}$$

• Why the downvote? Did I miss something? This one doesn't require partial fractions. – Jack's wasted life Oct 29 '16 at 6:59
• Shouldn't the upper and lower bounds for the integral be 0 & 1 respectively, and not vice-versa? – HeWhoMustBeNamed Oct 15 '17 at 7:19
• @Mr Reality $\frac{dx^{-1}}{dx}$ is negative, this minus signs allows us to make the upper bound 1 and lower bound 0. – Jack's wasted life Oct 18 '17 at 3:59
• Alright, thanks! – HeWhoMustBeNamed Oct 18 '17 at 8:44

$$\frac {x^3+3}{x^6(x^2+1)}=\frac{3}{x^6}-\frac{3}{x^4}+\frac{1}{x^3}+\frac{3}{ x^2}-\frac{1}{x}+\frac{x}{x^2+1}-\frac{3}{x^2+1}$$ So $$\int\frac {x^3+3}{x^6(x^2+1)}\,dx=-\frac{3}{5 x^5}+\frac{1}{x^3}-\frac{1}{2 x^2}+\frac{1}{2} \log \left(x^2+1\right)-\frac{3}{x}-\log (x)-3 \tan ^{-1}(x)$$ that is to say $$-\frac{3}{5 x^5}+\frac{1}{x^3}-\frac{1}{2 x^2}-\frac{3}{x}+\frac{1}{2} \log \left(\frac{x^2+1}{x^2}\right)-3 \tan ^{-1}(x)$$ Now, using the bounds, it seems to be quite fast.

• I accepted other answer as it didnt use partial fractions but yours is good too. – Archis Welankar Oct 29 '16 at 7:48
• @Claude Leibovici, I didn't get how you broke up the fraction in the first step. Can you tell me how you did that? – HeWhoMustBeNamed Oct 15 '17 at 2:39
• @MrReality. This is just partial fraction decomposition. – Claude Leibovici Oct 15 '17 at 3:05
• @Claude Leibovici Yes, I know that but I didn't get which techniques you followed. Is there someplace I can read up on it? – HeWhoMustBeNamed Oct 15 '17 at 3:11

First, break the integral in two

$$\int_0^{\infty} \frac{ x^3 + 3 }{x^6(x^2+1)} = \int_1^{\infty} \frac{dx}{x^3(x^2+1)} + 3 \int_1^{\infty} \frac{dx}{x^6(x^2+1)}$$

First, we integrat the second integral. Notice

$$\int\limits_1^{\infty} \frac{(1 +x^2 - x^2) dx }{x^6(x^2+1)} = \int\limits_1^{\infty} \frac{dx}{x^6} - \int\limits_1^{\infty} \frac{ d x}{x^4(x^2+1)} = \frac{1}{5} - \int\limits_1^{\infty} \frac{dx}{x^4} + \int\limits_1^{\infty} \frac{ dx }{x^2 ( x^2+1)} =$$

$$= \frac{1}{5} - \frac{1}{3} + \int\limits_1^{\infty} \left( \frac{1}{x^2} - \frac{1}{1+x^2} \right) dx = -\frac{2}{15} + \frac{ \pi }{2} +1$$

Now, for the first integral, use same trick

$$\int\limits_1^{\infty} \frac{ (1 + x^2 - x^2)dx}{x^3(x^2+1)} = \int_1^{\infty} \frac{dx}{x^3} - \int_1^{\infty} \frac{ dx }{x(x^2+1)} = \frac{1}{2} - \int_1^{\infty} \frac{ dx }{x(x^2+1)}$$

Using $x = \tan t$, we solve the last integral easily,

$$\int_1^{\infty} \frac{ \sec^2 t dt }{tan t \sec^2 t } = \int_1^{\infty} \frac{ \cos t dt }{\sin t} = \ln ( \sin t ) = \ln ( \frac{ x }{\sqrt{x^2+1}}) = \frac{1}{2} \ln \left( \frac{x^2 }{x^2+1} \right) \bigg|_1^{\infty} = \frac{1}{2} \ln (1/2)$$

Thus,

$$\int_0^{\infty} \frac{ x^3 + 3 }{x^6(x^2+1)} = \frac{-2}{15} + \frac{\pi}{2} + 1 + \frac{1}{2} - \frac{1}{2} \ln(1/2) = \boxed{ \frac{30 \ln 2+ 81 + 30 \pi }{60}}$$

• I have 3 doubts/questions regarding your answer: First, how did you get $\fracπ2$ in the first integral (I'm getting −$\fracπ4$). Second, that was a great factoring technique (I haven't seen anything like it)- you employed for breaking up the fractions- is there somewhere I could read up on it to learn the steps (are there steps or was it brute force... and what is this technique called?) Third- slight nitpicking- shouldn't it be 82 instead of 81 in the result? – HeWhoMustBeNamed Oct 23 '17 at 17:11