- #1
Natasha1
- 493
- 9
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
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