Integrate this indefinite integral: 1/(x-6)^2 dx

  • Thread starter Natasha1
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  • #1
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I need to integrate this indefinite integral: 1/(x-6)^2 dx

Here is my work...

Let u= x-6

du/dx=1

so: integral 1/u^2 du

= 3/u^3 + c (constant)
=3/(x-6)^3 + c

Have I gone wrong? And if so where? Thanks
 

Answers and Replies

  • #2
StatusX
Homework Helper
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Be careful. What is the derivative of 1/xn? So what is its integral?
 
  • #3
467
7
StatusX said:
Be careful. What is the derivative of 1/xn? So what is its integral?
not sure what the derivative is??? what is it?

The integral is n lnx so is this correct then

Here is my work...

Let u= x-6

du/dx=1

so: integral 1/u^2 du

= 2 lnu + c (constant)
= 2 ln(x-6) + c

Have I gone wrong? And if so where? Thanks
 
  • #4
StatusX
Homework Helper
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The derivative of 1/xn is -n/xn+1.
 
  • #5
VietDao29
Homework Helper
1,423
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Natasha1 said:
not sure what the derivative is??? what is it?

The integral is n lnx so is this correct then

Here is my work...

Let u= x-6

du/dx=1

so: integral 1/u^2 du

= 2 lnu + c (constant)
= 2 ln(x-6) + c

Have I gone wrong? And if so where? Thanks
You are almost there. Your last step is wrong...
Look again at your integral table, there should be something like this:
[tex]\int x ^ \alpha \ dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \ \alpha \neq -1[/tex]
[tex]\int \frac{1}{x} \ dx = \int x ^ {-1} \ dx = \ln (x) + C \quad \mbox{this is the case for } \alpha = -1[/tex].
And we also have:
[tex]\frac{1}{a ^ m} = a ^ {-m}[/tex]. Do you know this?
So you have:
[tex]\int \frac{du}{u ^ 2} = \int u ^ {-2} \ du = ?[/tex].
You can go from here, right? :smile:
 

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