Integration using substitution

[AdityaDev]

Homework Statement

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

Homework Equations

None

The Attempt at a Solution

I easily got the answer using partial fractions by splitting the integral as $\frac{Ax+B}{x^2+1}+\frac{C}{x}+\frac{D}{x^2}+\frac{E}{x^3}+...+\frac{H}{x^6}$ and then finding the coefficients by equating the numerator $x^2+3$ with the one obtained using partial fractions by taking LCM.
But is there an easier method?

 

[Mark44]

Homework Statement

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

Homework Equations

None

The Attempt at a Solution

I easily got the answer using partial fractions by splitting the integral as $\frac{Ax+B}{x^2+1}+\frac{C}{x}+\frac{D}{x^2}+\frac{E}{x^3}+...+\frac{H}{x^6}$ and then finding the coefficients by equating the numerator $x^2+3$ with the one obtained using partial fractions by taking LCM.
But is there an easier method?

The only "easier" method I can think of would be an ordinary substitution. Maybe there is one, but if there is, I don't see it. For this integral I would do exactly what you did.

 

[STEMucator]

Break up the integral:

$$\int \frac{x^2 + 3}{x^6(x^2 + 1)} dx = \int \frac{1}{x^4(x^2 + 1)} dx + \int \frac{3}{x^6(x^2 + 1)} dx$$

 

[SammyS]

Maybe a different split:
$\displaystyle\ \frac{x^2+3}{x^6(x^2+1)}=\frac{x^2+1}{x^6(x^2+1)}+\frac{2}{x^6(x^2+1)}=\frac{1}{x^6}+\frac{2}{x^6(x^2+1)}\$

Then noticing that the title of the thread is "Integration using substitution" and not "Integration by substitution" ... (maybe not intended but try it)

Simplify $\displaystyle\ \frac{2}{x^6(x^2+1)}\$ by letting $\ u=x^2\ .$ Just do this to aid in manipulating the rational expression.

This gives: $\displaystyle\ \frac{2}{u^3(u+1)}\$

Do partial fraction decomposition, or use the following reduction:
$\displaystyle\ \frac{1}{u^n(u+1)}=\frac{1}{u^n}-\frac{1}{u^{n-1}(u+1)}\$ (repeatedly).

Of course change $\ u\$ back to $\ x^2\$ before integrating.

 

[AdityaDev]

Maybe a different split:
$\displaystyle\ \frac{x^2+3}{x^6(x^2+1)}=\frac{x^2+1}{x^6(x^2+1)}+\frac{2}{x^6(x^2+1)}=\frac{1}{x^6}+\frac{2}{x^6(x^2+1)}\$

You can split it like this (better method): $$\frac{3x^2+3-2x^2}{x^6(x^2+1)}=\frac{3(x^2+1)-2x^2}{x^6(x^2+1)}=\frac{3(x^2+1)}{x^6(x^2+1)}-2\frac{x^2}{x^6(x^2+1)}=3\frac{1}{x^6}-2\frac{1}{x^4(x^2+1)}$$

 

[SammyS]

You can split it like this (better method): $$\frac{3x^2+3-2x^2}{x^6(x^2+1)}=\frac{3(x^2+1)-2x^2}{x^6(x^2+1)}=\frac{3(x^2+1)}{x^6(x^2+1)}-2\frac{x^2}{x^6(x^2+1)}=3\frac{1}{x^6}-2\frac{1}{x^4(x^2+1)}$$

I hesitate to put this up, but can't resist.
$\displaystyle\ \frac{3+x^2}{x^6(1+x^2)}=\frac{3+3x^2-2x^2}{x^6(1+x^2)}$

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

$\displaystyle\ =\frac{3(1+x^2)-2x^2(1+x^2)+2x^4(1+x^2)-2x^6}{x^6(1+x^2)}$​

Just a different way to get the partial fraction decomposition, I suppose.

 

[AdityaDev]

I hesitate to put this up, but can't resist.
$\displaystyle\ \frac{3+x^2}{x^6(1+x^2)}=\frac{3+3x^2-2x^2}{x^6(1+x^2)}$

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

$\displaystyle\ =\frac{3(1+x^2)-2x^2(1+x^2)+2x^4(1+x^2)-2x^6}{x^6(1+x^2)}$​

Just a different way to get the partial fraction decomposition, I suppose.

You get the the same thing if you use partial fractions like i did in post 0#

 

[SammyS]

You get the the same thing if you use partial fractions like i did in post #1

I would hope so.

The partial fraction method often seems rather tedious to me. I'm not sure this was any better.

Doing the substitution, u = x² like I suggest in post #4 above could make the partial fraction thing easier, especially after using the split you suggested.