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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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    How about long division/partial fractions? The original denominator can be factored...2012-11-15
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    I've seen the exact same problem on here quite recently. That and a harder problem...2012-11-15
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    @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?2012-11-15
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    Honestly, it would probably save both of us much time if you would go to wolfram alpha and have it "show steps"...2012-11-15
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    @melyong: good. +12012-11-15

2 Answers 2

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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 $\frac{1}{x^2+a}$.