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Find: $$\int \dfrac{x^4+2x-1}{(x^2+1)^3} dx$$

Now I have attempted to use partial fractions to split the integral into three different fractions, but this has not helped me integrate this function, the main problem being the quadratic factor of $x^2$ in the denominator while lacking linear factors of $x$ in the numerator. What should I do?

Any help is highly appreciated.

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  • $\begingroup$ Your title says that we must use partial fractions, but the main text doesn't. $\endgroup$
    – Kenny Lau
    Commented Jul 30, 2016 at 7:34
  • $\begingroup$ You don't have to, but it is the easiest way to do so. $\endgroup$
    – John Smith
    Commented Jul 30, 2016 at 7:43

1 Answer 1

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$$\dfrac{x^4-1+2x}{(x^2+1)^3}=\dfrac{x^2-1}{(x^2+1)^2}+\dfrac{2x}{(x^2+1)^3}$$

For the second integral choose $x^2+1=y$

$$\dfrac{x^2-1}{(x^2+1)^2}=\dfrac{1-\dfrac1{x^2}}{\left(x+\dfrac1x\right)^2}$$

$\displaystyle\int\left(1-\dfrac1{x^2}\right)dx=?$

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  • $\begingroup$ How did you know to split it like that? $\endgroup$
    – John Smith
    Commented Jul 30, 2016 at 7:40
  • $\begingroup$ @JohnSmith, Why? $$x^4+2x-1=(x^2-1)(x^2+1)+2x$$ $\endgroup$ Commented Jul 30, 2016 at 7:41

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