# Integral , just check if i'm right plz

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• Feb 8th 2009, 03:48 PM
Banned for attempted hacking
Integral , just check if i'm right plz
$\int (1+e^x)^{\frac{1}{2}} dx$

I did $u^2 = 1+e^x$

I got $2\int u^2/(u^2 -1)du$

Is this correct ?
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$\int x^3/(x-1)^4 dx$

I used parts $\int x^3(x-1)^-4 = (-1/3)(x^3/(x-1)^3 .....$

I know you could do u =x -1

But is my way wrong ?
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$\int 1/(x)(4x^2 +1)^{\frac{1}{2}}dx$
i did

u ^2 = (4x^2 +1)

I got the integral to

$\int 1/(u^2-1)du$

Is this right ?
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I just want you to tell me if i'm right because my book gives different answers when i do the integral my way.
• Feb 8th 2009, 04:50 PM
skeeter
Quote:

Originally Posted by ╔(σ_σ)╝
$\int (1+ex)^{\frac{1}{2}} dx$

why all the substitution?

$\int (1+ex)^{\frac{1}{2}} \, dx = \frac{2(1+ex)^{\frac{3}{2}}}{3e} + C$

did you mean ...

$\int (1+e^x)^{\frac{1}{2}} \, dx$ ?
• Feb 8th 2009, 04:52 PM
Banned for attempted hacking
Quote:

Originally Posted by skeeter
why all the substitution?

$\int (1+ex)^{\frac{1}{2}} \, dx = \frac{2(1+ex)^{\frac{3}{2}}}{3e} + C$

did you mean ...

$\int (1+e^x)^{\frac{1}{2}} \, dx$ ?

Yeah

If it was e^x , if it was ex then this would be elementary.
• Feb 8th 2009, 05:42 PM
elliotyang
Sorry i have to say, it look wrong at a glance.
Why you have to make all these subsitution? make some clear, easy and simple substitution
for sure your book will give you different answer because your method is incorrect.
• Feb 8th 2009, 05:52 PM
Banned for attempted hacking
How is my method wrong ?

I'm doing it the right way .

Okay if i'm wrong then what's the right way ?

I don't appreciate you telling me i did it wrong if you can't tell me the right thing to do. lol

u^2 = 1+e^x

is probably the most clear substitution.
• Feb 8th 2009, 06:15 PM
elliotyang
ok, well, u can use it to substitute.
u^2 = 1+e^x
2u du=e^x dx
du/dx=e^x/2u
integrate sqrt (1+e^x)dx= integrate u.e^x/2u du = 0.5 integrate (u^2-1) du
what you substitute is incorrect

sorry i do not know how to type all those notation
• Feb 8th 2009, 06:19 PM
elliotyang
perhaps you get confuse about the derivative
if you integrate f(x) dx
then you wont to substitute something to that
so it will become integrate f(u) dx . (du/dx) = f(u) du
• Feb 8th 2009, 06:23 PM
Banned for attempted hacking
Quote:

Originally Posted by elliotyang
ok, well, u can use it to substitute.
u^2 = 1+e^x
2u du=e^x dx
du/dx=e^x/2u
integrate sqrt (1+e^x)dx= integrate u.e^x/2u du = 0.5 integrate (u^2-1) du
what you substitute is incorrect

sorry i do not know how to type all those notation

So that means i'm right

because

u^2 = 1+e^x === > u^2 -1 = e^x right ?
u = sqrt(1 +e^x)

So what are you talking about ?

I think you're the confused one lol.

You don't always have to substitute something that is obvious .
• Feb 8th 2009, 06:26 PM
elliotyang
why don't you use my method to compute and see whether my answer is correct or not. since you are not believe what i have done so far.
• Feb 8th 2009, 06:38 PM
elliotyang
sorry, just now my mischief cousin playing with my laptop. your method are correct indeed. so sorry if he offended you.
the one you solve using integration is not encouraged. the working could be quite tedious. but u can free to use any substitution and method u like. as long as the answer is correct.
Wolfram Mathematica Online Integrator
this is online integrator, quite useful. you can always check the final answer is correct or not.

once again, sorry for what my cousin had done to u.
• Feb 8th 2009, 06:46 PM
Banned for attempted hacking
Mathematica sometimes spits out garbage like when i tried to integrate

$\int(tanx)^\frac{1}{2}$
• Feb 8th 2009, 06:53 PM
elliotyang
sometimes it does happen. perhaps the programming got certain bugs.
• Feb 8th 2009, 06:55 PM
o_O
Because it isn't a simple integral?

Let: $u^2 = \tan x \ \Leftrightarrow x = \arctan u^2 \ \Rightarrow \ dx = \frac{2u}{1+u^4} du$

So: $\int \sqrt{\tan x} \ dx \ = \ \int \frac{2u^2}{1+u^4} \ du$

And then you get some crazy suggestion.
• Feb 8th 2009, 06:58 PM
Banned for attempted hacking
Quote:

Originally Posted by o_O
Because it isn't a simple integral?

Let: $u^2 = \tan x \ \Leftrightarrow x = \arctan u^2 \ \Rightarrow \ dx = \frac{2u}{1+u^4} du$

So: $\int \sqrt{\tan x} \ dx \ = \ \int \frac{2u^2}{1+u^4} \ du$

And then you get something crazy.

I already did this question before.

Adding 2u^2 and -2u^2 to the denominator, makes the problem easy .

Then after that you can split it into a partial fraction decompose.

$\int \frac{2u^2}{1+u^4} \ du$ =

$\int \frac{2u^2}{1+2u^2 - 2u^2 +u^4} \ du$

$\int \frac{2u^2}{(u^2+1)^2 - 2u^2 } \ du$

using the identity a^2 - b^2 = (a-b)(a+b)

$\int \frac{2u^2}{(u^2 - √2u +1)(u^2 +√2u+1) } \ du$
• Feb 8th 2009, 07:04 PM
Banned for attempted hacking
$\int \frac{2u^2}{(u^2 - sqrt(2)u +1)(u^2 +sqrt(2)u+1) } \ du$