Integral Problem

[gadav478]

Good afternoon!

I am struggling with an integral problem: Integrate 5x + 2/(3x^5).
I understand that you can integrate each terms by itself so...
5x winds up being (5/2)x^2, but I am having trouble with integrating 2/(3x^5).
Here is my final answer (which isn't in the back of the book):
(5/2)x^2 - (6/4)x^-4 + C.
The answer per the book is:
(5/2)x^2 - 1/(6x^4) + C.

The way I am going about integrating 2/(3x^5) is:
2/(3x^5) = (2)(3x^-5) = 6x^-5
After integration I get: (-6/4)x^-4

Is there something I am doing wrong? Thanks for your help in advance.

 

[Deleted member 4993]

Good afternoon!

I am struggling with an integral problem: Integrate 5x + 2/(3x^5).
I understand that you can integrate each terms by itself so...
5x winds up being (5/2)x^2, but I am having trouble with integrating 2/(3x^5).
Here is my final answer (which isn't in the back of the book):
(5/2)x^2 - (6/4)x^-4 + C.
The answer per the book is:
(5/2)x^2 - 1/(6x^4) + C.

The way I am going about integrating 2/(3x^5) is:
2/(3x^5) = (2)(3x^-5) = 6x^-5

How is that? It should be:

2/(3x^5) = 2/3 * x ^-5

Integrate to get:

2/3 * 1/(-5+1) * x ^(-5+1) + C

Now continue.....

After integration I get: (-6/4)x^-4

Is there something I am doing wrong? Thanks for your help in advance.

.

 

[DrPhil]

Good afternoon!

I am struggling with an integral problem: Integrate 5x + 2/(3x^5).
I understand that you can integrate each terms by itself so...
5x winds up being (5/2)x^2, but I am having trouble with integrating 2/(3x^5).
Here is my final answer (which isn't in the back of the book):
(5/2)x^2 - (6/4)x^-4 + C.
The answer per the book is:
(5/2)x^2 - 1/(6x^4) + C.

The way I am going about integrating 2/(3x^5) is:
2/(3x^5) = (2)(3x^-5) = 6x^-5.....X
After integration I get: (-6/4)x^-4

Is there something I am doing wrong? Thanks for your help in advance.

Your error is what you did with the "3". It has to stay in the denominator:
2/(3x^5) = (2/3)x^(-5)

 

[gadav478]

Got it. Thank you so much.