Question

how do i integrate (1+r)^-2 dr?

Answer 1

\[\int \frac{1}{(1+r)^2}dt\]?

Answer 2

yes

Answer 3

but the end is dr not dt

Answer 4

use the power rule backwards

Answer 5

how do i use the power rule backwards?

Answer 6

\[\int x^ndx=\frac{x^{n+1}}{n+1}\]

Answer 7

can x in that case be a function, like 1+r?

Answer 8

so
\[\int (1+r)^{-2}dr=\frac{(1+r)^{-1}}{-1}\]

Answer 9

better written as \(-\frac{1}{r+1}\)

Answer 10

yes , you can treat \(1+r\) as \(x\) because it is just \(1+r\) and not say \(1+2r\)

Answer 11

um how are 1+r and 1+2r different when it comes to using the powerrule for integrals?

Answer 12

\[\int\limits f^n \left( x \right)f'\left( x \right)dx=\frac{ f ^{n+1}\left( x \right) }{ x+1 }\]

Answer 13

i thought (1+r) was a function itself.

Answer 14

is there a backwards chain rule?

Answer 15

Let \(x=1+r\) and then \(dx = dr\)

Answer 16

okay, what is next?

Answer 17

\[
\int x^{-2}\;dx
\]This is a simple integral.

Answer 18

ty

Answer 19

i have -x^-1
do i replace x with 1+r before evaluating the defined intgeral?

Answer 20

sure