Homework Help: Basic integration

1. Jun 14, 2017

Tanishq Nandan

1. The problem statement, all variables and given/known data
Solve the following integral: [(x^2+3)/(x^8+x^6)] dx

2. Relevant 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

3. 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

2. Jun 14, 2017

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.

3. Jun 14, 2017

Tanishq Nandan

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) ??

4. Jun 14, 2017

Ray Vickson

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
$$\frac{3+t}{t^3+t^4} = \frac{3+t}{t^3(1+t)} = \frac{A + B t + C t^2}{t^3} +\frac{D}{1+t}.$$

Last edited: Jun 14, 2017
5. Jun 15, 2017

Tanishq Nandan

Yup,got the asnwer.Thanks!!