# Solve Integral: ∫ (5x^2 - 42x + 24)/(x^3 - 10x^2 + 12x + 72) dx

In summary, the conversation discusses different approaches to solving the integral ∫ (5x^2 - 42x + 24)/(x^3 - 10x^2 + 12x + 72) dx, with the individual suggesting a factorization method and the other suggesting splitting the fraction into three separate fractions. The conversation ends with a suggestion to follow the latter approach, as it results in easier integrals.

## Homework Statement

Q. Solve ∫ (5x^2 - 42x + 24)/(x^3 - 10x^2 + 12x + 72) dx

## Homework Equations

∫ x^n dx = ∫ (1/n+1)x^(n+1) + C
∫ 1/x dx = ln x + C

## The Attempt at a Solution

Hi everyone,

Here's what I've done so far:

(i) I factored the bottom equation into (x + 2)(x - 6)^2

and then separated the fraction 1/[(x + 2)(x - 6)^2] into a sum of three fractions of the form:
A/(x + 2) + B/(x - 6) + C/[ (x-6)^2 ]

and found that A = 1/64, B = -1/64, C = 1/8

(ii) The only way I can see to go from here is to split the top integral into three integrals:

(1/64) ∫ (5x^2 - 42x + 24)/(x + 2) dx + (-1/64) ∫ (5x^2 - 42x + 24)/(x - 6) dx + (1/8) ∫ (5x^2 - 42x + 24)/(x-6)^2 dx

and solve them all individually using substitution, letting u = x+2, x-6 and x-6 respectively.

The answer I get in this way is: (5/8)x - 21/4 +2ln(x+2) + 3ln(x-6) + 6/(x-6) + C, C = constant of integration,

which is incorrect, according to the WebWork site, but I can't find any algebraic errors.
So, I'm thinking perhaps what I do in step (ii) is not the best way to do it.

Has anyone any suggestions?

Thanks for any help!

You don't want to do it that way. You want to split (5x^2 - 42x + 24)/((x+2)*(x-6)^2) into three fractions of the form A/(x + 2) + B/(x - 6) + C/[ (x-6)^2 ] with A, B and C constant.

Last edited:
You have split your fractions correctly. However splitting $1/((x-6)^2(x+2))$ instead of the entire integrand, like Dick suggests, results in you having to integrate three integrals which are very sensitive to mistakes. I checked the expression you wrote down in (ii) in mathematica and it does give you the correct answer so you must have made a mistake integrating somewhere. However I suggest you do it Dick's way, it gives easy values for A, B and C (integers) and results in three very easy integrals.

Last edited:

## 1. What is an integral?

An integral is a mathematical concept that represents the area under a curve. It is used to calculate the total value of a function over a specific interval or range.

## 2. How do I solve an integral?

To solve an integral, you can use techniques such as substitution, integration by parts, or partial fractions. You can also use online calculators or computer software to solve integrals.

## 3. What is the process for solving this specific integral?

The process for solving this specific integral involves using partial fractions to break down the fraction into smaller, more manageable parts. Then, you can use the power rule for integration to integrate each term separately.

## 4. Can this integral be solved analytically?

Yes, this integral can be solved analytically using the techniques mentioned above. However, it may require some algebraic manipulation and may be time-consuming.

## 5. What are the applications of solving integrals?

Solving integrals has many practical applications in fields such as physics, engineering, economics, and statistics. It can be used to calculate areas, volumes, work, and other important quantities in these fields.