- #1

- 467

- 7

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

- Thread starter Natasha1
- Start date

- #1

- 467

- 7

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

- #2

StatusX

Homework Helper

- 2,564

- 1

Be careful. What is the derivative of 1/x^{n}? So what is its integral?

- #3

- 467

- 7

not sure what the derivative is??? what is it?StatusX said:Be careful. What is the derivative of 1/x^{n}? So what is its integral?

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

- 2,564

- 1

The derivative of 1/x^{n} is -n/x^{n+1}.

- #5

VietDao29

Homework Helper

- 1,423

- 2

You areNatasha1 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

Look again at your

[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?

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