Integrating [(x^2+3)/(x^8+x^6)] dx by Substitution

[Tanishq Nandan]

Homework Statement

Solve the following integral: [(x^2+3)/(x^8+x^6)] dx [/B]

Homework Equations

The question has also said to integrate by substitution(though other methods are welcome)
That would mean substituting an expression in x with a variable,say, 't' such that the integral comes of the form f(t)dt ,which is eaier to evaluate using standard results,and then we can replace the value of t

The Attempt at a Solution

Broke down the numerator (x^2+3) into 2 terms (x^2+1) and 2 and then,separated the two fractions.
The first term came out to be (1/x^6)dx ,which is easy to integrate,but the second expression is really problematic: [2/(x^8 + x^6)]dx
I tried multiplying certain powers of x both to the numerator and denominator,but that hasn't worked well so far.(thought of substituting trigo terms,but that doesn't look promising either)
So..stuck

 

[andrewkirk]

Try working on the denominator rather than the numerator.

Use partial fraction decomposition to express the fraction as the sum
$$\frac{something}{x^6} + \frac{something\ else}{x^2+1}$$

The integration should then be easy.

 

[Tanishq Nandan]

Try working on the denominator rather than the numerator.

Use partial fraction decomposition to express the fraction as the sum
$$\frac{something}{x^6} + \frac{something\ else}{x^2+1}$$

The integration should then be easy.

Something=3
Something else=-2/x^4

The first term is easy enough,but then a similar problem is coming with the second term.
How to integrate
-2/(x^4+x^6) ??

 

[Ray Vickson]

Something=3
Something else=-2/x^4

The first term is easy enough,but then a similar problem is coming with the second term.
How to integrate
-2/(x^4+x^6) ??

Convert properly to partial fractions:
$$\frac{-2}{x^4+x^6} = \frac{-2}{x^4(1+x^2)} = \frac{A}{x^2} + \frac{B}{x^4}+\frac{C}{1+x^2}.$$
Alternatively, let $t = x^2$ and convert
$$\frac{-2}{t^2(1+t)}$$
to partial fractions, then put back $t = x^2$ later.

Also, you could have saved yourself a lot of trouble by doing it correctly from the start. Putting $x^2 = t$ in your $f(x) = (3+x^2)/(x^6+x^8)$ gives
$$\frac{3+t}{t^3+t^4} = \frac{3+t}{t^3(1+t)} = \frac{A + B t + C t^2}{t^3} +\frac{D}{1+t}.$$

 

[Tanishq Nandan]

Yup,got the asnwer.Thanks!