Question

\[\int\limits {3 \over (1+2x)^2}dx\]

Answer 1

sometimes i wonder what your grade level is...trigonometry then suddenly calculus lol =)) anyway..why not try to let u= 1 + 2x then do a u-substitution? you know how to do that right?

Answer 2

Umm. Mixture of grades... Um why can't you do it like this? I tried this, but got the wrong answer.\[3 \int\limits (1+2x)^{-2}\]\[3 \int\limits (1+2x)^{-1} \times 2\]??

Answer 3

you can...but you also have to multiply the integral by 1/2 so you maintain equalit...you multiplied by 2 to get a perfect integral..then you also pull a 1/2 outside the integral make sense?

Answer 4

Why do you bring a half outside?

Answer 5

let me try my first integral ...

\[\LARGE \int \frac{3}{(1+2x)^2}dx=\]
u=1+2x
2dx=du
dx=1/2du

\[\LARGE \int \frac{3}{u^2}\cdot \frac12du =\frac32\int u^{-2}du\]
am I right till here? lol hahahaha....

Answer 6

Answer is :\[-3/2(1+2x)^{-1}\]

Answer 7

Why times by half though?

Answer 8

u = (1 +2x)
du = 2dx

right? but you dont have a 2..so you multiply your integral by 2 to get a perfect integral...but this is no longer your original equation..you need to multiply by 1/2 because 1/2 times 2 is 1 so it's like nothing happened....but now you have a perfect integral in (1+2x)^-2 2dx right?

Answer 9

@lgbasallote is my integral right or not?

Answer 10

You need to times by half because if you don't, when you differentiate the result to check ,you have double the original expression.

Answer 11

it's right :D

Answer 12

aawww... I feel like a BOSS ahahah... lol

Answer 13

OK, can you write the steps out one by one (literally again)... not sure I completely understand... Sorry!!

Answer 14

let u = (1+ 2x)
du = 2dx

follow that so far?

Answer 15

here's why it's multiplied by 1/2

when \(\LARGE 1+2x=u\)
now we take derivative of1+2x
so we have 2 then we take derivative of "u" which is 1
so 2dx=1*du
2dx=du
then we multiply both sides by 2
dx=1/2 du (my version) ..

Answer 16

your original equation was

\(\LARGE \int \frac{3dx}{(1+2x)^2}\)

\(\LARGE 3\int \frac{dx}{(1+2x)^2}\)

i think you got the part up to here

Answer 17

Yeh

Answer 18

right now you cannot integrate that...but if your numerator had 2dx then it would be integrable by the general power formula

\(\LARGE 3\int \frac{2dx}{(1+2x)^2}\)

you got that right?

Answer 19

Yes

Answer 20

but that's not equal to your original equation! you added a 2!! but!! if it were..

\(\LARGE 3\int \frac{\frac{1}{2} 2dx}{(1+2x)^2}\)

your numerator is just equal to 1 right? so it's as if you didnt change anything...you get that?

Answer 21

Yeah

Answer 22

but you still need 2dx so it is a perfect integral...so take out the 1/2

\(\LARGE \frac{3}{2} \int \frac{2dx}{(1+2x)^2}\)

you get that?

Answer 23

yes

Answer 24

so now you can integrate it by

\(\LARGE \int (1+2x)^{-2} \)

power formula...

\(\LARGE \frac{(1+2x)^{-2 +1}}{-2 +1}\)

then multiply it by your 3/2

Answer 25

Ok. Thank you!!!

Answer 26

haha yeah its like the chain rule backwards...thats why its simpler to just use u substitution, less chance of making mistake

Answer 27

yeah..i hate it when teachers dont just teach that :/ it's confusing when you have more values