How to Expand Rational Expressions for Integration

[karush]

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

this was homework for a section of expanding rational expressions

$$\frac{x^4}{4+x^2}=x^2+\frac{16}{x^2 +4}-4$$

I don't see how W|F got this expansion?

 

[kaliprasad]

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

this was homework for a section of expanding rational expressions

$$\frac{x^4}{4+x^2}=x^2+\frac{16}{x^2 +4}-4$$

I don't see how W|F got this expansion?

This you can do by synthetic divison or as x^2+ 4 is in denominator

$x^4 = x^2(x^2+4) - 4x^2$
$= x^2 +(x^2+4) -4(x^2 + 4) + 16$
$= (x^2-4)(x^2+4) + 16$

deviding both sides by $x^2+4$ on both sides you get the result

 

[Prove It]

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

this was homework for a section of expanding rational expressions

$$\frac{x^4}{4+x^2}=x^2+\frac{16}{x^2 +4}-4$$

I don't see how W|F got this expansion?

$\displaystyle \begin{align*} \frac{x^4}{x^2 + 4} &= \frac{x^4 + 4x^2 - 4x^2}{x^2 + 4} \\ &= \frac{x^4 + 4x^2}{x^2 + 4} - \frac{4x^2}{x^2 + 4} \\ &= \frac{x^2 \left( x^2 + 4 \right) }{x^2 + 4} - \frac{4x^2}{x^2 + 4} \\ &= x^2 - \frac{4x^2}{x^2 + 4} \\ &= x^2 - \frac{4x^2 + 16 - 16}{x^2 + 4} \\ &= x^2 - \frac{4x^2 + 16}{x^2 + 4} - \left( \frac{-16}{x^2 + 4} \right) \\ &= x^2 - \frac{4 \left( x^2 + 4 \right) }{x^2 + 4} + \frac{16}{x^2 + 4} \\ &= x^2 - 4 + \frac{16}{x^2 + 4} \end{align*}$

 

[karush]

I like that, so now we have

$$\int x^2 \ dx + \int\frac{16}{x^2 + 4} \ dx- \int 4 \ dx $$

$$\frac{{x}^{3}}{3}+8\arctan\left({\frac{x}{2}}\right)+4x +C$$

I hope

 

[Prove It]

I like that, so now we have

$$\int x^2 \ dx + \int\frac{16}{x^2 + 4} \ dx- \int 4 \ dx $$

$$\frac{{x}^{3}}{3}+8\arctan\left({\frac{x}{2}}\right)+4x +C$$

I hope

I don't know why you put + instead of - for the sign of "4x". Everything else is correct.

 

[karush]

Got it..
$$\frac{{x}^{3}}{3}+8\arctan\left({\frac{x}{2}}\right)-4x +C$$