Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $

Question

Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$

I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.

Answer 1

hint: $\dfrac{x^2+3}{x^6(x^2+1)}= x^{-6}+\dfrac{2}{x^6(x^2+1)}$. At this point, let $x = \dfrac{1}{y}$, then the second part becomes $\dfrac{y^6}{y^2+1}= y^4-\dfrac{y^4}{y^2+1}=y^4-y^2 + \dfrac{y^2}{y^2+1}=y^4-y^2+1 - \dfrac{1}{y^2+1}$,and you can take it from here.

Answer 2

That rational expression, like all rational expressions, can be broken into partial fractions. You may be making the common mistake of trying to change the expression to

$$\frac{x^2+3}{x^6(x^2+1)}=\frac{A}{x^6}+\frac{Bx+C}{x^2+1}$$

However, you need to include all the lesser powers of the irreducible factors. In other words, you want

$$\frac{x^2+3}{x^6(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x^5}+\frac{F}{x^6}+\frac{Gx+H}{x^2+1}$$

Try that, then come back if you cannot finish.

Answer 3

Both methods work just fine here:

(1) Partial fractions One can apply the Method of Partial Fractions to any rational expression, and since the denominator of the integrand is already factored into a product of irreducible polynomials (over $\Bbb R$), we can proceed directly to writing the decomposition: $$\frac{x^2 + 3}{x^6 (x^2 + 1)} = \frac{A}{x^6} + \frac{B}{x^5} + \frac{C}{x^4} + \frac{D}{x^3} + \frac{E}{x^2} + \frac{F}{x} + \frac{G}{x^2 + 1} .$$

In fact, since the integrand is even, only even terms appear in the decomposition: $$\frac{x^2 + 3}{x^6 (x^2 + 1)} = \frac{A}{x^6} + \frac{C}{x^4} + \frac{E}{x^2} + \frac{G}{x^2 + 1} .$$

(2) Trigonometric substitution This is probably less efficient than partial fractions here, but it's not a bad option: The usual substitution $x = \tan \theta, dx = \sec^2 \theta\, d\theta$, transforms the integral to $$\int \frac{\tan^2 \theta + 3}{\tan^6 \theta},$$ which can be handled in several ways. One approach is to write the numerator as $\sec^2 \theta + 2$ and splitting the integral as $$\int \frac{\sec^2 \theta \,d\theta}{\tan^6 \theta} + 2\int \cot^4 \theta \,d\theta .$$ The first can be handled with substitution and the second with the usual trick of writing $\cot^2 = \csc^2 - 1$ and applying the elementary integral $\int \csc^2 \theta = - \cot \theta + C$.

Answer 4

Let's try

$$\frac{x^2+3}{x^6(x^2+1)}=\frac ax+\frac b{x^2}+\frac c{x^3}+\frac d{x^4}+\frac e{x^5}+\frac f{x^6}+\frac{gx+h}{x^2+1}\implies$$

$$x^2+3=ax^5(x^2+1)+bx^4(x^2+1)+cx^3(x^2+1)+dx^2(x^2+1)+ex(x^2+1)+f(x^2+1)+(gx+h)x^6$$

and now:

$$x=0\implies 3=f\;,\;\;coef. \;x^7\implies 0=a+g\;,\;\;coef.\;x^6\implies 0=b+h$$

$$coef.\;x^5\implies 0=a+c\;,\;\;coef.\;x^4\implies0=b+d\;,\;\;coef.\;x^3\implies 0=c+e$$

$$coef.\;x^2\implies1=d+f\;,\;\;coef.\;x\implies0=e\;$$

From here:

$$f=3\;,\;\;d=-2\;,\;\;b=2\;,\;\;h=-2\;,\;\;c=e=a=g=\implies0$$

$$\frac{x^2+3}{x^6(x^2+1)}=\frac2{x^2}-\frac2{x^4}+\frac3{x^6}-\frac2{x^2+1}\implies$$

$$\int\frac{x^2+3}{x^6(x^2+1)}dx=-\frac2x+\frac2{3x^3}-\frac3{5x^5}-2\arctan x+C$$

Answer 5

Since $$ \begin{align} \frac{x^2+3}{x^6(x^2+1)} &=\frac1{x^6}+\frac2{x^6(x^2+1)}\\ &=\frac1{x^6}+\frac{2(x^6+1)}{x^6(x^2+1)}-\frac2{x^2+1}\\ &=\frac1{x^6}+\frac{2(x^4-x^2+1)}{x^6}-\frac2{x^2+1}\\ &=\frac3{x^6}-\frac2{x^4}+\frac2{x^2}-\frac2{x^2+1} \end{align} $$ we have $$ \int\frac{x^2+3}{x^6(x^2+1)}\,\mathrm{d}x =C-\frac3{5x^5}+\frac2{3x^3}-\frac2x-2\arctan(x) $$

Answer 6

Below is a nice general way of computing the partial fraction that is a generalization of Heaviside's cover up method to nonlinear denominator factors.

$$\begin{eqnarray} \frac{x^2+3}{(x^2+1)x^6} &=&\ \ \frac{ax+b}{x^2+1}+ \frac{f(x)}{x^6}\quad\text{so clearing denominators yields}\\ x^2+3\ &=&\ \ (ax+b)x^6\! + (x^2+1) f(x)\quad\text{so evaluaing at $x=i$ then $x = -i$} \\ 2\ &=& -\!ai - b \\ 2\ &=& \ \ \ \ ai - b\quad\text{so subtracting the prior two equations yields}\\ 0\ &=& -\!2ai\end{eqnarray}$$

Therefore $\ a=0\ $ so $\ 2 = -b\ $ so $\ b = -2\ $ and, finally, we can compute $\,f(x)\,$ by substituting these $\,a,b\,$ values into the first equation. Then the integral of the decomposition is easy.

See here for further discussion of this method - which often greatly simplifies calculations.

Answer 7

I learned something weird in attempting this integral. When you see $x^2+a^2$, the four substitutions that come to mind are $x=a\sec\theta$, $x=a\csc\theta$, $x=a\sinh\theta$, and $x=a\,\text{csch}\,\theta$. The first $3$ leave a big power of a trigonometric or hyperbolic function in the denominator, so I tried $x=\text{csch}\theta$. $$\begin{align}\int\frac{x^2+3}{x^6(x^2+1)}dx&=-\int\frac{(\text{csch}^2\theta+3)\text{csch}\theta\coth\theta}{\text{csch}^2\theta\coth^2\theta}d\theta\\ &=-\int\frac{(1+3\sinh^2\theta)\sinh^4\theta}{\cosh\theta}d\theta\\ &=\int\left(-3\cosh\theta\sinh^4\theta+2\cosh\theta\sinh^2\theta-2\cosh\theta+2\text{sech}\theta\right)d\theta\end{align}$$ Now, the first $3$ terms above are no problem but the last one has the issue that the other $3$ secants have a common form for their indefinite integrals, $$\int\sec\theta d\theta=\ln|\sec\theta+\tan\theta|+C$$ $$\int\csc\theta d\theta=-\ln|\csc\theta+\cot\theta|+C$$ $$\int\text{csch}\,\theta d\theta=-\ln|\text{csch}\,\theta+\coth\theta|+C$$ But $\text{sech}\,\theta$ doesn't fit this pattern so I can never remember its integral, so I always have to substitute $z=\tanh\left(\frac{\theta}2\right)$ and end up with $$\int2\text{sech}\,\theta\,d\theta=4\tan^{-1}\left(\tanh\frac{\theta}2\right)+C=4\tan^{-1}\left(\frac1{x+\sqrt{x^2+1}}\right)+C$$ Now, not only does this look horrible, but it exposes the integrator for having used the unimaginative 'Weierstrass' substitution. This can be fixed up because $$\begin{align}4\tan^{-1}\left(\frac1{x+\sqrt{x^2+1}}\right)&=2\tan^{-1}\left(\frac{\frac2{x+\sqrt{x^2+1}}}{1-\frac1{(x+\sqrt{x^2+1})^2}}\right)\\ &=2\tan^{-1}\left(\frac{2(x+\sqrt{x^2+1})}{x^2+2x\sqrt{x^2+1}+x^2}\right)\\ &=2\tan^{-1}\frac1x=\pi-2\tan^{-1}x\end{align}$$ So that means that $$\int2\text{sech}\theta\,d\theta=-2\tan^{-1}x+C$$ But that in turn implies that $$\int\text{sech}\,\theta\,d\theta=\tan^{-1}\sinh\theta+C_1=-\tan^{-1}\text{csch}\,\theta+C_2$$ OK, so I will have to remember that. Differentiation confirms this identity. Maybe it would be easier to remember the forms $$\int\text{csch}\,\theta\,d\theta=-\coth^{-1}\cosh\theta+C$$ $$\int\sec\theta\,d\theta=\tanh^{-1}\sin\theta+C$$ $$\int\csc\theta\,d\theta=-\tanh^{-1}\cos\theta+C$$ Nahh... Getting back to the task at hand, $$\begin{align}\int\frac{x^2+3}{x^6(x^2+1)}dx&=-\frac35\sinh^5\theta+\frac23\sinh^3\theta-2\sinh\theta-2\tan^{-1}\text{csch}\theta+C\\ &=-\frac3{5x^5}+\frac2{3x^3}-\frac2x-2\tan^{-1}x+C\end{align}$$