Integration by Partial Fractions Help

In summary, to solve the integral of [x^3+4]/[x^2+4] dx, you can use long division to rewrite it as x+(-4x+4)/[x^(2)+4]. Then, using the rule ax^2+bx+c ⇒ (Ax+B)/(ax^2+bx+c), you can rewrite it as (Ax+B)/[x^(2)+4] and solve for A and B. To integrate the remaining fraction, split it into two fractions and use u-substitution on one of them. The other fraction is a well-known integral.
  • #1
StrangeCharm
23
12

Homework Statement


∫ [x^(3)+4] / [x^(2)+4] dx

Homework Equations


N/A

The Attempt at a Solution


I know that the fraction is improper, so I used long division to rewrite it as x+(-4x+4)/[x^(2)+4].
Given the form S(x)+R(x)/Q(x), Q(x) is a distinct irreducible quadratic factor [x^(2)+4].
I used the rule ax^2+bx+c ⇒ (Ax+B)/(ax^2+bx+c) to rewrite it as (Ax+B)/[x^(2)+4].
I then solved for A and B and got A=-4 and B=4.
I am now trying to solve ∫ [ x + (-4x+4)/(x^(2)+4) ] dx
I know that ∫x=(1/2)x^2, but I am stuck with integrating (-4x+4)/[x^(2)+4].
(I tried u-substitution and that didn't work.)
 
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  • #2
StrangeCharm said:

Homework Statement


∫ [x^(3)+4] / [x^(2)+4] dx

Homework Equations


N/A

The Attempt at a Solution


I know that the fraction is improper, so I used long division to rewrite it as x+(-4x+4)/[x^(2)+4].
Given the form S(x)+R(x)/Q(x), Q(x) is a distinct irreducible quadratic factor [x^(2)+4].
I used the rule ax^2+bx+c ⇒ (Ax+B)/(ax^2+bx+c) to rewrite it as (Ax+B)/[x^(2)+4].
I then solved for A and B and got A=-4 and B=4.
I am now trying to solve ∫ [ x + (-4x+4)/(x^(2)+4) ] dx
I know that ∫x=(1/2)x^2, but I am stuck with integrating (-4x+4)/[x^(2)+4].
(I tried u-substitution and that didn't work.)
Hello StrangeCharm. Welcome to PF !

Split ##\displaystyle \frac{-4x+4}{x^2+4}## into two fractions:

##\displaystyle \frac{-4x}{x^2+4} + \frac{4}{x^2+4}##

The integral of one of them can be done via u-substitution. The other is a fairly well known integral.
 
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  • #3
SammyS said:
Split ##\displaystyle \frac{-4x+4}{x^2+4}## into two fractions:

##\displaystyle \frac{-4x}{x^2+4} + \frac{4}{x^2+4}##

The integral of one of them can be done via u-substitution. The other is a fairly well known integral.
Thanks for the tip!
 
  • #4
I'm almost done with the problem but am having trouble with integrating 4/[x^(2)+4]. It looks like integrating 1/(1+x^2)dx and getting arctan(x); however, I'm not sure how to simplify the fraction into a similar form.
Also, I'm wondering whether it is valid to make ∫4/(4+x^2)dx ⇒ ∫[2/(2+x)]^2dx and using u-substitution with u=x+2. I'm not sure because doing this would get a different result from the method above.
 
  • #5
StrangeCharm said:
I'm almost done with the problem but am having trouble with integrating 4/[x^(2)+4]. It looks like integrating 1/(1+x^2)dx and getting arctan(x); however, I'm not sure how to simplify the fraction into a similar form.
Also, I'm wondering whether it is valid to make ∫4/(4+x^2)dx ⇒ ∫[2/(2+x)]^2dx and using u-substitution with u=x+2. I'm not sure because doing this would get a different result from the method above.
Is (2 + x)2 = 4 + x2 ? No.

Use u substitution: u = 2x .
 
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  • #6
Okay, right... I just realized that. Thanks for the help!
 

Related to Integration by Partial Fractions Help

1. What is integration by partial fractions?

Integration by partial fractions is a method used to simplify and solve certain types of integrals involving rational functions. It involves breaking down a complex rational function into simpler fractions that can be easily integrated.

2. When is integration by partial fractions used?

Integration by partial fractions is used when an integral involves a rational function, which is a polynomial divided by another polynomial. It is especially useful when the degree of the numerator is less than the degree of the denominator.

3. How do you solve integrals using partial fractions?

To solve an integral using partial fractions, you first need to factor the denominator of the rational function. Then, you set up a system of equations using the coefficients of each term in the numerator and equate them to the coefficients of the corresponding terms in the denominator. This system of equations can be solved to find the unknown constants, which can then be used to rewrite the original rational function as a sum of simpler fractions. These fractions can then be integrated individually.

4. What are the benefits of using integration by partial fractions?

One of the main benefits of using integration by partial fractions is that it can simplify complex integrals and make them easier to solve. It also allows for the use of other integration techniques, such as substitution, which may not have been possible with the original rational function.

5. Are there any limitations to integration by partial fractions?

Integration by partial fractions can only be used for certain types of integrals involving rational functions. It may not work for all integrals and may not always be the most efficient method of integration. In some cases, it may also be difficult to solve the system of equations to find the unknown constants.

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