Question

integrate 1/ [sqrt x^2(1+sqrt x)^2] dx

Answer 1

\[\int \frac{1}{\sqrt{x^2} (1+\sqrt x)^2}dx\]

Is that really the question?

Answer 2

yes that it. got it for an assignment

Answer 3

Your professor hates you...

Answer 4

lol, i think so!

Answer 5

\[ \int \frac{1}{\sqrt{x^2} (1+\sqrt x)^2}dx = \int \frac{1}{x (1+\sqrt x)^2}dx\]

Answer 6

i'm attempting to follow how you got your answer @foolformath

Answer 7

start by letting
u = sqrt x
du = 1/(2 sqrt x) dx
That gets us to:
\[2\int\frac{1}{u(u+1)^2}\]
From there we need to do partial fractions...

Also, sqrt x^2 doesnt equal x. That works only if x is a positive number, which we cannot assume

Answer 8

Indeed agreene is right ...... I am just being a fool :P

Answer 9

x^2 will always be positive?

Answer 10

no matter what x is

Answer 11

lol thats a very common mistake... gotta remember those plus/minuses :P

So, yeah, you need to do partial fractions, then you'll have to do I think two more substitutions and you'll end up with a big algebraic soup...

Answer 12

so technically what fool said is correct however you'll get +- and that will just be a disaster

Answer 13

thanks so much!

Answer 14

@taya
http://www.wolframalpha.com/input/?i=1%2F+%5Bsqrt+x%5E2%281%2Bsqrt+x%29%5E2%5D+dx

You can click show steps on the right--it gets really messy real quick >.<

Answer 15

yeah the square roots works as follows . sign :

$$ \sqrt[n]{a^n} = a \text{ if $n$ is odd } $$

$$ \sqrt[n]{a^n} = |a| \text{ if $n$ is even } $$

Answer 16

i tried wolframalpha and got even more confused, but I will have another look @ it

Answer 17

and Outkast... the issue is the slope of the eqn when you assume that.

|x| = sqrt x^2 is a true statement if even
x = sqrt x^2 iff x is odd

Answer 18

don't use wolfram alpha for indefinite integrals even simple things loooks messy ...

Answer 19

I just realized I went insane when typing that definition for you outkast... look at fool's he did it correctly.

Answer 20

ooh, okay... i will look @ fools, thanks

Answer 21

it's actually a lot easier if you substitute x = u^2 because it reduces to:

2 * integral of ( du/(u(1+u)) )

Answer 22

= ln|x| - 2ln|1 + sqrt(x)| + C

Answer 23

ooh, that is so simple that way thank you @james