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Could I have some help for this question?

$$\int \frac {x^4}{x^4 +5x^2 +4} \, dx$$

I've reduced the equation to...

$$\int \frac {1}{1 + 5x^{-2} + 4x^{-4}} \, dx$$

But I'm stuck after this step. Could I have some help for this question?

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4  
How about long division/partial fractions? The original denominator can be factored... –  The Chaz 2.0 Nov 15 '12 at 5:35
    
I've seen the exact same problem on here quite recently. That and a harder problem... –  Mike Nov 15 '12 at 5:46
    
@TheChaz, I've tried using partial fraction. But I could not get the numerator for the partial fractions. $(x^4 + 5x^2 = 4) = (x^2+\frac 52)^2 -(\frac 32)^2; x^4 = 4A-B +x^2 (A+B)$. How should I proceed from here? –  melyong Nov 15 '12 at 5:51
    
Honestly, it would probably save both of us much time if you would go to wolfram alpha and have it "show steps"... –  The Chaz 2.0 Nov 15 '12 at 7:30
    
@melyong: good. +1 –  Babak S. Nov 15 '12 at 10:15

2 Answers 2

Hint:

Using partial fraction, you can write the integrand as

$$\frac {x^4}{x^4 +5x^2 +4}= 1+\frac{1}{3(x^2+1)} - \frac{16}{3(x^2+4)}.$$

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Using the fact that $x^4+5x^2+4=(x^2+1)(x^2+4)$, change the initial integrand to a form only involving $\frac{1}{x^2+a}$.

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