How Do You Integrate the Rational Function ∫((x^2+x-7)/(x+3))dx?

[whatlifeforme]

Homework Statement

Integrate the following:

Homework Equations

∫(x^2+x-7)/(x+3)

The Attempt at a Solution

The only way I can think of solving this would be to split up each term into a separate fraction.

 

[Dick]

Homework Statement

Integrate the following:

Homework Equations

∫(x^2+x-7)/(x+3)

The Attempt at a Solution

The only way I can think of solving this would be to split up each term into a separate fraction.

It's easier if you do the polynomial division. Divide (x^2+x-7) by (x+3).

 

[whatlifeforme]

how would i do that? i tried to separate the numerator into a product of two terms (x+3)(another term)... but that doesn't work.

 

[Dick]

how would i do that? i tried to separate the numerator into a product of two terms (x+3)(another term)... but that doesn't work.

No, factoring doesn't work. Hopefully you learned polynomial division at some point and then forgot you knew it. See http://en.wikipedia.org/wiki/Polynomial_long_division

 

[Ray Vickson]

how would i do that? i tried to separate the numerator into a product of two terms (x+3)(another term)... but that doesn't work.

Why not substitute u = x+3 and do a u-integration instead?

 

[Hysteria X]

Why not substitute u = x+3 and do a u-integration instead?

Wouldn't that be a little too long as the numerator is not directly divisible by x+3?? or is there any shortcuts

 

[whatlifeforme]

so i would have:

1. ∫(x+3)(x-2) - 1 / (x+3)
2. ∫(x-2) - ∫1/(x+3) [i'm not sure how to take this integral though]

 

[Hysteria X]

so i would have:

1. ∫(x+3)(x-2) - 1 / (x+3)
2. ∫(x-2) - ∫1/(x+3)

Yep that should work :)

 

[Whovian]

[i'm not sure how to take this integral though]

Use a substitution of u=x+3, du=dx. A general rule of thumb is that if L is a linear function of x, then integrating

$$\int\left(f\left(L\left(x\right)\right)\right) \cdot\mathrm{d}x$$

is a matter of making a simple substitution. Another thing I've learned is that when integrating functions, try making any substitutions which might simplify it and see where they get you, unless you're almost certain they won't get you anywhere.

 

[Dick]

so i would have:

1. ∫(x+3)(x-2) - 1 / (x+3)
2. ∫(x-2) - ∫1/(x+3) [i'm not sure how to take this integral though]

Same advice as to Hysteria X. Write ((x+3)(x-2)-1)/(x+3). That's what you really meant. That's the polynomial division part alright. The first integral is easy, the second is a substitution. Think 'log' for the actual integration.

 

[whatlifeforme]

i have for the answer:

x^2 - 2x - ln(x+3) + c

 

[Dick]

i have for the answer:

x^2 - 2x - ln(x+3) + c

I think the x^2 term is missing something.

 

[SammyS]

Why not substitute u = x+3 and do a u-integration instead?

For this integration, Ray's suggestion appears to be most straight forward. (I realize that in general it's important to know how to do long division when integrating a rational expression for which the degree of the numerator is equal to or greater than hte degree of the denominator.)

$\displaystyle \int \frac{x^2+x-7}{x+3}\,dx$

Let u = x+3  → x = u-3  → dx = du  → x^2+x-7 = (u-3)^2+(u-3)-7 = u^2 - 5u - 1

The integral becomes $\displaystyle\ \ \int \frac{ u^2 - 5u - 1<br /> }{u}\,du = \int (u-5-u^{-1})\,du$