# Evaluate the integral $\int \frac {1}{x^4 +1} dx$( finding the constants in the partial fraction method).

Evaluate the integral $$\int \frac {1}{x^4 +1} dx$$.

This question is answered here :

But I do not know how they found the constants in the partial fraction method, could anyone explain this for me please?

• Note that $x^4+1=(x^2+1)^2-2x^2=(x^2+1)^2-(\sqrt{2}x)^2$. – Batominovski Nov 19 '18 at 10:16
• – Dave L. Renfro Nov 19 '18 at 13:55

## 2 Answers

If you are doing the integral from $$0$$ to $$\infty$$ it's worth noting the following. Consider the integral $$\frac{1}{2\pi i} \int \frac{\log z}{1+z^n} dz, \qquad n=4$$ around the classic keyhole contour, that is to say, start at 0 just above the real axis, go along the real axis to $$\infty$$, then anticlockwise in a circle until you get to $$\infty$$ just below the real axis, then back to zero. The log funciton is different on the two sides of the branch cut. The rest is residue calculus. Note that this works for any $$n\ge 2$$, and emerges as $$(\pi/n) \mathrm{cosec} (\pi/n)$$.

A related trick is $$\frac{1}{2\pi i} \int \frac{z^a}{1+z^n} dz$$ around the same contour, and let $$a\to0$$ whereupon the branch cut disappears(!) but the answer remains intact.

• Does this answer the question? – Szeto Nov 19 '18 at 12:26
• Well it does if you are tasked with doing the integral from $0$ to $\infty$ and mistakenly assume that the first thing to do must be to figure out the indefinite integral, and then stick the limits in. – Richard Martin Nov 19 '18 at 12:30
• Which, I hasten to add, is exactly what I would have done when I was twelve! – Richard Martin Nov 19 '18 at 13:16

You can also do this: When you have factorised the term $$x^4+1$$ into factors $$p(x) & q(x)$$ then $$\frac{1}{x^4+1} = \frac {Ax+M}{p(x)}+\frac{Bx+N}{q(x)}$$ and then to find A, B, M and N just take values of x for both sides and equate to find them just like linear equations in two variables.

• I am not speaking about the general method ..... in this problem the numbers are difficult ...... choosing the value of x is difficult – hopefully Nov 19 '18 at 9:46
• One of easiest value putting is x=0 and if you can see x=√2 also helps! Also if you see the IIT-JEE papers you will find these types of problem to be solved in less than 5 minutes and the method taught to the aspirants is the same method. It is just the issue of finding so called special x value so that equation is simpler. And for that my friend practice is the best method. – jayant98 Nov 19 '18 at 9:52
• I have tried those values and they do not solve my problem also – hopefully Nov 19 '18 at 9:56
• $x=0$ implies $1=M+N$, $x = \sqrt(2)$ implies $1 = Ax+M + 5Bx+5N$, using $-\sqrt(2)$ analogue $1 = -5Ax+5M -Bx+N$, i.e. $Ax+5Bx = -5Ax-Bx$, i.e. $A+B=0$. Now get a handy 4th equation ... – Stockfish Nov 19 '18 at 10:49