Question

Can someone help me anti-differentiate the equation (2/(1-4x)^1/2) without using substitution?

Answer 1

\[\int\limits^{?}2\div \sqrt{1-4x}\] it's an indefinite integral

Answer 2

it's much easier to solve if you use u-sub. Why not use it?

Answer 3

The directions specifically say not too, if I could I would know how to do it, haha.

Answer 4

Could you use the equation editor tool to write down the integral. It might help to clarify things.

Answer 5

All I can think of is to multiply by the conjugate.

Answer 6

I believe it is:
\( \displaystyle \int \frac{2}{\sqrt{1 -4x}} \; \text{d}x \)

Answer 7

I'd say rewrite it as 2(1-4x)^(-1/2), then find the integral of (1-4x)^(-1/2). After that find what factor you need, and basically you are finished.

Answer 8

Well, find the integral of (1-4x)^(-1/2) by starting out with C(1-4x)^(1/2) and readjusting C so it fits wat you want. Basically guessing game lol

Answer 9

You are technically still using u-sub, but it's more like you are using the chain rule. It should be allowed though.

Answer 10

\[\int\frac{2}{\sqrt{1-4x}}dx=\frac{2}{-4}\int\frac{-4}{\sqrt{1-4x}}dx=-\frac{1}{2}\int\sqrt{1-4x}(-4dx)\]

Answer 11

By inspection...

Answer 12

You pretty much have to use substitution though, because it was necessary to use the chain rule to get it as a derivative.

Answer 13

@wio , I'm guessing his textbook/learning material wants him to find antiderivatives by guessing, and this is how my textbook introduced me to antiderivatives. But assuming the power rule, and making a initial guess and scaling it with a constant (usage of the chain rules) is using the chain rule and not u-sub

Answer 14

Lolz. Use chain rule. First off you know the final result would have the (1-4x). SO we put that in first and then add 1 to the power. We will leave the 2 outside of the integral sign since it will take care of itself in the end.
\[\huge 2\int\limits (1-4x)^{-\frac{1}{2}}= (1-4x)^{-\frac{1}{2}+1}\]
After that when you differentiate using the chain rule, you would bring the power down the front.
\[\huge =\frac{1}{2}(1-4x)^{\frac{1}{2}}\]
But we don't have a half in the front before, so we have to cancel hat by multiplying by 2.
We end up having this:
\[\huge =2[2(1-4x)^{\frac{1}{2}}]\]
\[\huge =4(1-4x)^{\frac{1}{2}}\]

Answer 15

@Azteck THATS WHAT I SAID BUT WITH BIG INTEGRAL SIGNY :O

Answer 16

cancel that*

Answer 17

No, I prefer to cancel hats >:(

Answer 18

@alaskamath , get what we are saying?

Answer 19

+C.

Answer 20

Forgot the constant. o dear me. SOrry about that.

Answer 21

\[\huge=4(1-4x)^{\frac{1}{2}}+C\]
If you want teachers to think you're cool at maths I suggest you convert that exponent/power into a square root sign. Makes it "neater".
\[\huge=4\sqrt{1-4x} + C\]

Answer 22

Thank you guys sooooo much :) I got it now.

Answer 23

Give azteck medal ;)

Answer 24

Maybe one other alternative, if you like making yourself suffer, is to use extended binomial theorem and expand out the 2(1 - 4x)^(-1/2) into a series. If you leave that in sigma notation, it'd be integrating the interior of the sum w.r.t. x. Once you integrate that (it'd be a simple power rule), you have to convert it back into a closed form. Not very fun imo, but it could work. :p

Answer 25

@Azteck

But we don't have a half in the front before, so we have to cancel hat by multiplying by 2.
We end up having this:
=2[2(1−4x)^1/2]

Why would it be \[2[2\sqrt{1-4x}]
instead of
[2[.5(\sqrt{1-4x} )]\]?

Answer 26

Sorry about that. Typo. \[2[\frac{1}{2}(1-4x)^{\frac{1}{2}}]\]
\[=\sqrt{1-4x}

Answer 27

\[=\sqrt{1-4x}\]

Answer 28

My computer gets lag every time you write it in big font.

Answer 29

Sorry about that.

Answer 30

Okay, that makes more sense :) Thank you again!

Answer 31

And don't forget the constant.

Answer 32

No worries.