Question

Integral question

Answer 1

\[\int\limits \dfrac{x^2-1}{2x^3}~dx\]

Answer 2

@SolomonZelman

Would I change it to \((x^2 - 1)*(2x^{-3})\)?

Answer 3

\(\color{black}{\displaystyle \int \frac{2x(x^2-1)}{4(x^2)(x^2)}dx}\)

Answer 4

Then, \(\color{black}{\displaystyle z=x^2}\). (I am tired of using u.)

Answer 5

Hmm that's a clever approach :d hehe

Answer 6

You see what I'm doing though?
I basically multiplied top and bottom by \(2x\), and now I can do a you sub.

Answer 7

I'm confused on how you did that...

Answer 8

\(\color{black}{\displaystyle \frac{\color{red}{(}x^2-1\color{red}{)\times 2x}}{2x^3\color{red}{\times 2x}} =\frac{2x(x^2-1)}{4(x^2)(x^2)}}\)

Answer 9

agree ?

Answer 10

Okay, I see that now

Answer 11

\(\color{black}{\displaystyle \int \frac{2x(x^2-1)}{4(x^2)^2}dx}\) and now do the sub.

Answer 12

but that is stupid, lets restart

Answer 13

\(\color{black}{\displaystyle \int \frac{x^2-1}{2x^3}dx}\)

\(\color{black}{\displaystyle \int \frac{x^2}{2x^3}-\frac{1}{2x^3}dx}\)

\(\color{black}{\displaystyle \int 2x^{-1}-2x^{-3}dx}\)

Answer 14

I have over-thought, and over-thought by far.
Sorry on that:)

Answer 15

Ohh, I forgot about splitting the fraction

Answer 16

yes, and so did I:)

Answer 17

Well, you got it from here :)

Answer 18

Wait, my instructor said the exponent can't be -1 because that would make the integral to the power of 0?

Answer 19

WHAT ??

Answer 20

\(\color{black}{\displaystyle \int \frac{1}{x} dx = \ln|x|+C}\)

Answer 21

Would it be ln|2x|?

Answer 22

Final answer: \(ln|2x| + \dfrac 1{x^2} + c\)

Answer 23

Actually, and technically (since C is arbitrary) and knowing the rules of logarithms ln|2x|+C = ln|2|+ln|x|+C= ln|x|+(ln|2|+C) = ln|x|+C, so it would not be incorrect, BUT

\(\color{black}{\displaystyle \int 2x^{-1}-2x^{-3}dx}\)

this is what you are really dealing with.
\(\color{black}{\displaystyle 2\int x^{-1}dx-2\int x^{-3}dx}\)

Answer 24

after computing the integrals individually, you get,

\(\color{black}{\displaystyle 2(\ln|x|+C_1)-2(-\frac{3}{2} x^{-2}+C_2)}\)

\(\color{black}{\displaystyle 2\ln|x|+2C_1+3x^{-2}-2C_2)}\)

\(\color{black}{\displaystyle 2\ln|x|+3x^{-2}+C}\)

((remember, those constants are arbitrary, and even if you add non-arbitrary (fixed) constants to them, or other arbitrary constants, you will just result in another arbitrary constant "C"))

Answer 25

Where did the -3/2 come from?

Answer 26

\(\large \color{black}{\displaystyle \int \frac{2}{x^3}dx= \int 2x^{-3}dx=2\cdot \frac{1}{-3\color{red}{+1}}}\cdot x^{-3\color{red}{+1}}\)

Answer 27

oh ...

Answer 28

I accounted for the 2 outside the second integral twice.
Thanks for catching me.

Answer 29

Couldn't you just do 2ln|x| - \(\dfrac{2x^{-2}}{-2}\)and the -2 cancels?

Answer 30

Yes,

Answer 31

Okay :)

Answer 32

and of course +C.

Answer 33

Yep :)

Answer 34

Or, your grade for that integral is 0.
(Because out of infinitely many possible answers, you only wrote one possible answer, as few of the people I know like to say.)

Answer 35

that is if you didn't add +C .... well, I think we got it.

Answer 36

Okay haha
I have 2 more indefinite integrals... Would you mind helping me with the last 2 of these?

Answer 37

yes.

Answer 38

and btw, that my overthing would be for the case where instead of x^2-1, you would have Sqrt(x^2-1).

Answer 39

overthinking **

Answer 40

anyway ... go ahead, and in any post you like.

Answer 41

Gotcha, I will go open a new question for the next one