Question

Help please!

Consider the integral integral of ((x^-2)-4)/x^3

A) Solve this integral by using u-substitution

B) Solve the integral another way, using algebra to simplify the integrand first.

C) How must your two answers be related? Use algebra to prove this relationship.

Answer 1

did you try u = x^-2 -4 ?

Answer 2

du = ... ?

Answer 3

so du = -2x^-3 dx

Answer 4

yes
du = -2/x^3 dx

so, du/(-2) = dx/x^3
plug this in your original integral where you see dx/x^3

Answer 5

so then that would be integral of u/(-2)du

Answer 6

thats correct, now evaluate the integral

Answer 7

so then it would be... ((x^-2)-4)/-2?

Answer 8

naah, integral of u is actually u^2/2 :)

Answer 9

wait...ummmm...how? :(

Answer 10

\(\Large \int x^n dx = \dfrac{x^{n+1}}{n+1}+c\)

Answer 11

but where does the bottom 2 come from? :(

Answer 12

u has the exponent =n =1

denominator = n+1 = 1+1 =2

Answer 13

oh because -2^1 is still -2

Answer 14

yeah

Answer 15

but you have u^2

so,
\(\Large (u^{-2}-4)^2/2+c\)
would be your answer for 1st

Answer 16

i mean
\(\Large (x^{-2}-4)^2/2+c\)

Answer 17

how come it's not -2?

Answer 18

oops, -2 in the denominator, yes ...

Answer 19

\(\Large -(x^{-2}-4)^2/2+c\)

Answer 20

ok cool so it is integral of u/-2du and becomes u^2/-2 and the du disappears

Answer 21

leaving -(x^-2-4)/2 + c

Answer 22

so how would i solve for that algebraically?

Answer 23

did i miss the 2 ???
let me write the steps, and ask if you have any doubts in that

\(u = x^{-2}-4 \\ du/(-2) = dx/x^3 \\ \int u/(-2)du = (-1/2)\times u^2/2+c = (-u^2/4)+c \\ \Large = [(x^{-2})-4]^2/4+c \)

Answer 24

wait how does 4 get into the equation? :(

Answer 25

the 4 in the denominator ??


-1/2 times u^2/2 = ...

Answer 26

for the denominator i mean

Answer 27

and i missed the negative sign
\(u = x^{-2}-4 \\ du/(-2) = dx/x^3 \\ \int u/(-2)du = (-1/2)\times u^2/2+c = (-u^2/4)+c \\ \Large =- [(x^{-2})-4]^2/4+c\)

Answer 28

i thought the equation for the integral was u/-2du

Answer 29

it was

Answer 30

for integral of u, we get u^2/2

and that -2 was already in the denominator
so, -2 times 2 = -4

Answer 31

sorry i still don't get where the 2 came from :(

Answer 32

the 2 in u^2/2 ??

Answer 33

so i have this...

u= x^(-2)-4

du = -2x^(-3) dx

-1/2du=x^(-3)

Answer 34

yeah :(

Answer 35

and (-1/2)du=1/(x^3)dx and that becomes...

integral of u/(-2)du

Answer 36

so where does the other 2 come from :(

Answer 37

\(\Large \int x^n dx = \dfrac{x^{n+1}}{n+1}+c \\ \Large \int x^1 dx = \dfrac{x^{1+1}}{1+1}+c\)

Answer 38

\(\Large (-1/2)\int u^1 dx = (-1/2)\dfrac{u^{1+1}}{1+1}+c = \dfrac{-u^2}{4}+c\)

Answer 39

oh! i get it now :D

Answer 40

so how would that be solved algebraically?

Answer 41

i am glad you got it :)
that was your A) part.

now lets do B) part , then we will algebraically simplify it...

Answer 42

right ?
can you write each term in the form of x^n ?