Question

find (x+1/x)^2 dx

Answer 1

are you to integrate?

Answer 2

yes.

Answer 3

try first

Answer 4

I tried and I couldn't get, if not I wouldn't have posted here.

Answer 5

show me what you did

Answer 6

(x^2 + x+x/4)
(x^3/3 +x^2/2+x^-3)

Answer 7

\[(x+\frac{ 1 }{ x })^{2}dx\]
is that the given expression?

Answer 8

first,
is that [(x+1)/x]^2 or [(x+ (1/x)]^2 ?

Answer 9

as what @shinalcantara said.

Answer 10

try to expand the expression first then

Answer 11

I already did.

Answer 12

so what did you get?

Answer 13

I just need someone to explain to me how to get the workings and correct me, as I have posted my workings.

Answer 14

\[
\int (x^2+1/x^2+2) dx
\]

Answer 15

1/x^2 what did you get it from?

Answer 16

:(

Answer 17

Huh?

Answer 18

\[
(x + \frac{1}{x})~(x+\frac{1}{x}) = x \times x + \frac{x}{x} + \frac{x}{x} + \frac{1}{x} \times \frac{1}{x} = x^2+2+\frac{1}{x^2} \\
OR \\
\sf \huge ~ x^2 +\frac{1}{x^2} +2
\]

Answer 19

I got it before you replied me. thanks anyways.

Answer 20

\[
\int x^2~dx+ \int \frac{1}{x^2}~dx + 2 \int ~dx
\]

Answer 21

can you take it from where I left off?

Answer 22

remember that:
\[
\int \frac{1}{x^2} dx = \int x^{-2} dx
\]