How to Expand Rational Expressions for Integration

In summary, for the integration of $\frac{x^4}{4+x^2}$, you can use synthetic division or divide by the denominator to get $x^2 - 4 + \frac{16}{x^2 + 4}$, which can be further broken down into $x^2 - 4 + \frac{16}{x^2 + 4} = x^2 - \frac{4x^2 + 16}{x^2 + 4} + \frac{16}{x^2 + 4}$. This can be integrated as $\int x^2 \ dx + \int\frac{16}{x^2 + 4} \ dx- \
  • #1
karush
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$$\int\frac{x^4}{4+x^2}dx$$

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?
 
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  • #2
karush said:
$$\int\frac{x^4}{4+x^2}dx$$

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
 
  • #3
karush said:
$$\int\frac{x^4}{4+x^2}dx$$

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?

$\displaystyle \begin{align*} \frac{x^4}{x^2 + 4} &= \frac{x^4 + 4x^2 - 4x^2}{x^2 + 4} \\ &= \frac{x^4 + 4x^2}{x^2 + 4} - \frac{4x^2}{x^2 + 4} \\ &= \frac{x^2 \left( x^2 + 4 \right) }{x^2 + 4} - \frac{4x^2}{x^2 + 4} \\ &= x^2 - \frac{4x^2}{x^2 + 4} \\ &= x^2 - \frac{4x^2 + 16 - 16}{x^2 + 4} \\ &= x^2 - \frac{4x^2 + 16}{x^2 + 4} - \left( \frac{-16}{x^2 + 4} \right) \\ &= x^2 - \frac{4 \left( x^2 + 4 \right) }{x^2 + 4} + \frac{16}{x^2 + 4} \\ &= x^2 - 4 + \frac{16}{x^2 + 4} \end{align*}$
 
  • #4
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
 
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  • #5
karush said:
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.
 
  • #6
Got it..
$$\frac{{x}^{3}}{3}+8\arctan\left({\frac{x}{2}}\right)-4x +C$$
 

FAQ: How to Expand Rational Expressions for Integration

What is the definition of the integral of a rational function?

The integral of a rational function is a mathematical operation that calculates the area under the curve of a rational function. It is denoted by ∫f(x)dx and is a fundamental part of calculus.

How is the integral of a rational function calculated?

The integral of a rational function is calculated using the technique of integration, which involves finding the antiderivative of the function and evaluating it at the upper and lower limits of integration. This process is also known as the fundamental theorem of calculus.

Can the integral of a rational function be negative?

Yes, the integral of a rational function can be negative. This occurs when the area under the curve is below the x-axis, resulting in a negative value for the integral.

Are there any special cases when integrating a rational function?

Yes, there are a few special cases when integrating a rational function. One such case is when the function has a vertical asymptote, in which case the integral may be evaluated using the method of partial fractions. Another special case is when the function has a non-constant numerator, in which case the integral may be evaluated using the method of substitution.

What are the applications of the integral of a rational function?

The integral of a rational function has many real-world applications, particularly in physics and engineering. It is used to calculate the area under a velocity-time curve to determine displacement, the area under a force-distance curve to calculate work, and the area under a concentration-time curve to determine the amount of a substance in a solution.

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