Question

When taking the anti-derivative of a definite integral, if the quantity of the integral/function = "36(2x + 1)^(-3)dx", does it become...
1) (36x)(2x + 1)^(-2)
Or...
2) 36(2x + 1)^(-2)
Or...
3) (36x)(2x + 1)^(-2)(2x^2 + x)

??

Any and all help is greatly appreciated! :)

Answer 1

try "u sub"

Answer 2

... So you can't evaluate this by taking an anti derivative and plugging in the limits?

Answer 3

you can use advanced guessing

Answer 4

\(\large \int 36(2x + 1)^{-3}dx\)


since it is of x^n form, you should get something like below :

\(\large 36\frac{(2x + 1)^{-3+1}}{-3+1} \)

Answer 5

you need to fix it a bit, cuz its not just x in the parenthesis,
its 2x+1 !

Answer 6

Okay, so you don't add an exponent to the 36, and your don't use the chain rule (that would be backwards). Okay...

Answer 7

holdup - you still need to fix it, yes you need to use chain rule backwards

Answer 8

Fix what? Which post?

Answer 9

\(\large \int 36(2x + 1)^{-3}dx\)

since it is of x^n form, you should get something like below :

\(\large 36\frac{(2x + 1)^{-3+1}}{-3+1} \)

since its (2x+1) inside the parenthesis, divide above stuff by (2x+1)' = 2

\(\large 36\frac{(2x + 1)^{-3+1}}{2(-3+1)} + c \)