Question

\[\int\limits_{?}^{?}(x+1/x)^2dx\]

Answer 1

u can open the square , and then do long division

Answer 2

If you open the square, you have \[\int\limits_{}^{}x^2 +2 +1/x^2+k\]
I just forgot how to integrate...

Answer 3

is it:
\[\int\limits\left(\frac{x+1}{x}\right) ^{2} \]

Answer 4

u know wut long divison is

Answer 5

No.. \[\int\limits_{}^{}[(x) + (1/x)]^2\]

Answer 6

no when u simplify it wud b {(1+(1/x)}^2

Answer 7

x^3/3 + 2x - 1/x +c

Answer 8

i hate sumone pops out from middle jus telling answer, i was making her understand!

Answer 9

just convert (x+1/x)^2 to x^2+ 1/x^2 +2 and use the integration rules

Answer 10

Why is the second to the last one -1/x?

Answer 11

arnab she doesnt know long division!

Answer 12

because 1/x^2 = x^-2
when u integrate x^n, it will be x^(n+1)/(n+1)

Answer 13

@ order, u dont know the formula of (a+b)^2??

Answer 14

Um I think so... how does x^-2 become -1x?

Answer 15

I mean -1/x

Answer 16

integrate x^-2
so, it will be x^-1/(-1)=-1/x

Answer 17

\[\int\limits \left(x+\frac{1}{x}\right) ^{2} = \int\limits x^{2}+\frac{1}{x^{2}} +2 \]
\[ \frac{x^{3}}{3} +x^-2 + 2x \]
\[\frac{x^{3}}{3} +\frac{x^{-1}}{-1} +2x + C\]

Answer 18

Ok, that makes a lot more sense :) Thanks Mimi, and also you Arnab... I just couldn't understand when you wrote it...

Answer 19

woops typo:
\[ \frac{x^{3}}{3} +x^{-2} + 2x \]

Answer 20

welcome :)