Question

Calculus?

find the indefinite integral and check the result by differentiation. integrate ((x^2+2x-3) / x^4) dx

Answer 1

divide, integrate term by term using the power rule backwards

Answer 2

that is, integrate
1 / x^2 + 2 / x^3 - 3 / x^4 one at a time

Answer 3

I leave 1 alone right?

Answer 4

what 1?

Answer 5

the 1 in the numerator

Answer 6

rewrite in exponential notation as
x^(-2) + 2x^(-3) -3 x^(-4) and go from there

Answer 7

oh ok

Answer 8

clear or no? you are going to use
integral x^n = x^(n+1))/(n+1) for each term

Answer 9

x^(-2+1)/(-2+1)

Answer 10

thats for the first term yes?

Answer 11

yes, better known as - 1/x

Answer 12

2x^(-3+1)/(-3+1)

Answer 13

do I simplify that one too?

Answer 14

of course, you need to compute them all
that one is - 1/ x^2

Answer 15

how are you computig them? could you please explain?
And -3x^(-4+1)/(-4+1) = ?

Answer 16

VBG: The following may be clearer. I've used satellite73's suggestion that you "divide, then integrate term by term."\[\int\limits_{-}^{-}(\frac{ 1 }{ x ^{2} }+\frac{ 2 }{ x ^{3} }-\frac{ 3 }{ x ^{2} })dx\]

Answer 17

Hint: rewirte 1/(x^2) as x^(-2) and use the power rule for integration.

Answer 18

Mathmale ... you lost me... also it's hard for me to understand all the" [\int\limits_{-}^{-}(\frac{ 1 }{ x ^{2} }+\frac{ 2 }{ x ^{3} }-\frac{ 3 }{ x ^{2} })dx\] "

Answer 19

VBG: sorry. I don't know how to get Equation Editor to print the integral operator without limits of integration, so I've written - and - as upper and lower limits. Ignore that completely.

Rules of exponents tells us that 1/x^2 = x^(-2).

Integrating that, using the power rule for integration, results in (x^(-1)) / (-1), or -x^(-1), or simply -1/x.

OK with this so far?

Answer 20

\[\int\limits_{}^{}\frac{ 1 }{ x ^{2} }dx=\int\limits_{}^{}x ^{-2}dx=\frac{ -1 }{ x }+C\]

Answer 21

i think the equation button or like isn't working today

Answer 22

In that case, please try the Draw utility. Hope that works!

Answer 23

ok I think I understand up to there

Answer 24

VBG: IDAREYOU's comment makes me wonder whether you actually received the expression I'd tried to write to you. Sorry if Equation Editor let me and you both down.

Answer 25

hmmm I think it did.

Answer 26

Is it still possible for you to help me? I really need it

Answer 27

@eliassaab please help?

Answer 28

@ganeshie8 @amoodarya

Answer 29

http://www.twiddla.com/1469274
cme here

Answer 30

VBG: I'd suggest you re-post your latest question. I'm available for a while and would try to help you.

Answer 31

hey mathmale. It's ok ganesh got this one!