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

1. Jan 25, 2006

### Natasha1

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

2. Jan 25, 2006

### StatusX

Be careful. What is the derivative of 1/xn? So what is its integral?

3. Jan 25, 2006

### Natasha1

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. Jan 25, 2006

### StatusX

The derivative of 1/xn is -n/xn+1.

5. Jan 25, 2006

### VietDao29

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