Because if you use the change of variable
y = 1 + x
then x = y - 1 and dy = dx.
Thus
\int{x(1 + x)^{-5}\,dx} = \int{(y - 1)y^{-5}\,dy}
= \int{y^{-4} - y^{-5}\,dy} after expanding the brackets.
I found some integral problems online and have been using them to practice. I'm trying to integrate x(x + 1)^-5 with intervals of infinity and zero.
The solution here given seems to take advantage of a shortcut. That integral is changed to the integral of y^-4 dy minus the integral of y^-5 dy, both with boundaries of infinity and 1, instead of infinity and zero.
When I check the integral at Wolfram Alpha I get the integral as being (4x + 1) / 12 (x + 1)^4
Which ties out to the same answer using the shortcut of the y^-4 minus y^-5 with the changed interval.
Can someone walk through integrating x(x + 1)^-5 to get to the (4x+1) / 12(x+1)^4,
and then also explain why the shortcut works of taking y^-4 - y^-5?
I've attached the solution file, which shows the shortcut.
Have a nice day!
Chris
Here's an image, until LaTeX comes back on.
By plugging the following code into this page (mostly done with search-and-replace)
Code:\\ \text{Because if you use the change of variable} \\ \\y = 1 + x \\ \\\text{then }x = y - 1\text{ and }dy = dx\text{.} \\ \\ \\\text{Thus} \\ \\\int{x(1 + x)^{-5}\,dx} = \int{(y - 1)y^{-5}\,dy} \\ \\ = \int{y^{-4} - y^5\,dy}\text{ after expanding the brackets.}