Am I integrating this right: (x^2 + 3x + 11)/(x+2)^4 ?

[Lo.Lee.Ta.]

Am I integrating this right: (x^2 + 3x + 11)/(x+2)^4 ?

1. ∫(x² + 3x + 4)/(x+2)⁴dx


2. With these sorts of problems, I think about integration by partial fractions.

So in this case, the denominator factors are all the same, so I have to make each one with different exponents.
I wrote:

A/(x + 2) + B/(x + 2)² + C/(x + 2)³ + D/(x + 2)⁴

I multiplied to get common denominators:

A(x + 2)³ + B(x + 2)² + C(x + 2) + D = x² + 3x + 11

Now I need to figure out what A, B, C, and D equal.

When I substitute x=-2, D=9

A(x + 2)³ + B(x + 2)² + C(x + 2) + 9 = x³ + 3x + 11

I think I am supposed to take the derivative now to figure out the other values.

2x + 3 = 3A(x + 2)² + 2B(x + 2) + C

When I substitute x=-2, C= -1

Now, am I able to take the derivative again?
That's what I did next:
Substituting x=-2,

2 = 6A(x + 2) + 2B
B=1

If x=1,

2 = 6A(x + 2) + 2(1)
A=0

...Am I even doing this right? :/
I don't want to go on if I'm not even doing this part right!
Thank you very much! :D

 

[Dick]

1. ∫(x² + 3x + 4)/(x+2)⁴dx


2. With these sorts of problems, I think about integration by partial fractions.

So in this case, the denominator factors are all the same, so I have to make each one with different exponents.
I wrote:

A/(x + 2) + B/(x + 2)² + C/(x + 2)³ + D/(x + 2)⁴

I multiplied to get common denominators:

A(x + 2)³ + B(x + 2)² + C(x + 2) + D = x² + 3x + 11

Now I need to figure out what A, B, C, and D equal.

When I substitute x=-2, D=9

A(x + 2)³ + B(x + 2)² + C(x + 2) + 9 = x³ + 3x + 11

I think I am supposed to take the derivative now to figure out the other values.

2x + 3 = 3A(x + 2)² + 2B(x + 2) + C

When I substitute x=-2, C= -1

Now, am I able to take the derivative again?
That's what I did next:
Substituting x=-2,

2 = 6A(x + 2) + 2B
B=1

If x=1,

2 = 6A(x + 2) + 2(1)
A=0

...Am I even doing this right? :/
I don't want to go on if I'm not even doing this part right!
Thank you very much! :D

If your original integrand was (x^2+3x+11)/(x+2)^4 (you probably have a typo), then, yes, you got the partial fractions right.

 

[Lo.Lee.Ta.]

Oh, yes, was a typo! Thanks for letting me know it's right so far! :)

 

[Lo.Lee.Ta.]

Alright, now I have:

∫(0/x+2) + (1/(x+2)²) - 1/(x+2)³ + 9/(x+2)⁴dx

U-substitution!

u= x+2
du=dx

∫u^-2 - u^-3 + 9u^-4du

= -u^-1 - u^-2/-2 + 9u^-3/-3

= -1/(x+2) + 1/(2(x+2)²) - 2/(x+2)^-3

...Is this right?
Thanks so much! :)

 

[SammyS]

Alright, now I have:

∫(0/(x+2)) + (1/(x+2)²) - 1/(x+2)³ + 9/(x+2)⁴dx

U-substitution!

u= x+2
du=dx

∫u^-2 - u^-3 + 9u^-4du

= -u^-1 - u^-2/-2 + 9u^-3/-3

= -1/(x+2) + 1/(2(x+2)²) - 2/(x+2)^-3

...Is this right?
Thanks so much! :)

You can check by taking the derivative of the result.

By the way: This problem can be done more easily by using u-substitution at the outset, with u = x+2 .