Question

what is the integral of ( x^2+2/X^2-1)dx?

Answer 1

Well, what's the integral of x^2 ?
and of 2/x^2
and of -1?

Answer 2

So what's the integral of x^2 dx ?

Answer 3

x^3/3

Answer 4

right. So now integrate the other two terms.

Answer 5

the 2x?

Answer 6

integral 2/x^2 dx
and integral -1 dx

Answer 7

*i don't understand. what is the integral of x^2/x^2-1?

Answer 8

am i going to use u substitution?

Answer 9

No.

What is the integral of x^(-2) dx ?

Answer 10

uhm. x^(-1)/-1?

Answer 11

Yes, that's right. Or just -1/x.

We are evaluating the integral one term at a time:\[\int\limits ( x^2+2/x^2-1) \ dx = \int\limits x^2 \ dx + \int\limits 2x^{-2} \ dx + \int\limits -1 \ dx \]

Answer 12

We now have the first two terms. Can you take it from here?

Answer 13

ah. i see. thanks. kamsahamnida!

Answer 14

uhm. wait. why did it become integral of 2x^-2?

Answer 15

because 1/x^2 = x^(-2)

Answer 16

yes but it also has +1 doesn't it matter? so you mean the all the terms in the denominator goes to the numerator like (x^2-1)^-1?

Answer 17

oh, your integral is this:
\[\int\limits \left( x^2 + \frac{2}{x^2 - 1} \right) \ dx\]
?

Answer 18

You've got to use brackets otherwise it's not clear.

Answer 19

uhm no. the one you wrote first was my integral..

Answer 20

Ok. So in this case you want to write 2/(x^2 - 1) as partial fractions

2/(x^2 - 1) = A/(x-1) + B/(x+1)

Now find A & B.

And the reason we do that is the terms on the right-hand side are easy to integrate.

Answer 21

So it is this:
\[\int\limits (x^2 + 2/x^2 - 1) \ dx\]

Answer 22

Yes?

Answer 23

yes.

Answer 24

So ... integrate terms by term
x^2
2x^(-2) and
-1

Answer 25

why is it 2x^(-2)?

Answer 26

because\[\frac{2}{x^2} = 2x^{-2}\]

Answer 27

because 1/a = a^(-1)

Answer 28

e.g., 10^(-2) = 1/100

Answer 29

yes i get that but did the x^2 came from the denominator beacuse the x^2 in the denominator still has -1

Answer 30

\[\frac{2}{x^2} = 2 \frac{1}{x^2} = 2 (x^2)^{-1} = 2 x^{2(-1)} = 2x^{-2}\]

Answer 31

1/x = x^(-1)
1/x^2 = x^(-2)
1/x^n = x^(-n)

Answer 32

can we do that directly? didn't the \[x ^{2}\] have -1?

Answer 33

The point is
\[\int\limits \frac{2}{x^2} \ dx = - \frac{2}{x} \ \ (+C)\]
however you arrive at that result.

Answer 34

thanks. uhm last question how do you integrate (x^2/x^2-1)

Answer 35

\[\frac{x^2}{x^2 - 1} = 1 + \frac{1}{x^2 -1} = 1 + \frac{A}{x-1} + \frac{B}{x+1}\]

This is called a partial fraction decomposition. You need now to find the A and B which make this equation correct and then integrate every term on the RHS.

Answer 36

OMG.! that looks like its difficult.

Answer 37

wait. my integral was [(x^2+2)/(x^2-1)]dx so its the same with ∫(x2+2/x2−1) dx. the answer is −2x (+C)?

Answer 38

Then
\[\frac{x^2 + 2}{x^2 - 1} = 1 + \frac{3}{x^2 - 1}\]
Now use the partial fractions technique above and if you don't know it, go to your text book. It's really not so bad, but it requires familiarity.