Question

Find the indefinite integral of
integral of [(-x)/( (x+1) - (sqrt of x+1)]dx

Answer 1

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

Answer 2

does the equation make sense teh way i wrote it

Answer 3

its an indefinite integral not a definite one so tahts y there is still ? on teh top n side

Answer 4

\[\int\frac{-x}{x+1}dx\ - \int \sqrt{x+1}dx\] this?

Answer 5

no u see how u separated the squareroot of x+1....that is supposed to b included in teh denominator...so its teh integral of
num: -x and den: (x+1) - (sqrt of x+1)

Answer 6

i mean how did u do write it liek a fraction..wen i do it its always s slash sign

Answer 7

\[\int \frac{-x dx}{(x+1) - \sqrt{x+1}}\] this?

Answer 8

yes exxcept the dx is on the side of teh whole equation instead of in teh numerator...does taht make a dffnrce

Answer 9

it does not

Answer 10

ok so what would you do first?

Answer 11

well we are supposed to see if any of the der look like a derivative of anotehr so we can set taht to u

Answer 12

u-substitution..wil it help here?

Answer 13

you tell me, can you do a u sub here? or can you do some algebra to manipulate the problem?

Answer 14

i guess algebra since i dont see how any of them looks like a derivative of another

Answer 15

good, so what algebra?

Answer 16

not sure

Answer 17

but see the instructions say to use u-sub though but idk how to do that??

Answer 18

im not sure how alegbra helps

Answer 19

yes u sub will come after algebra. ill give you a hint, you can multiply the top and bottom by the same thing and it will be like multiplying the whole thing by 1

Answer 20

u mean mltiply it by its conjugate

Answer 21

like multiply the top by (x+1) +(sqrt of x+1)

Answer 22

is taht wat u mean?

Answer 23

yes, very good, top and bottom remember.

Answer 24

ok let me wrk this out...

Answer 25

so the top will be \[(-x^2 -x) + -x \sqrt{x+1}\]

Answer 26

for the denominator does it \[(x^2+2x+1) - (x+1) = x^2 +x\]

Answer 27

ok so combine everything

Answer 28

integral is \int

Answer 29

\[(-x^2-2x \sqrt{x+1}) / (x^2+x)\] dx

Answer 30

the top wont combine like that retry

Answer 31

just dont combine it like you did and try breaking the fraction up into pieces that you think can be integrated

Answer 32

hmm im so leave teh numerator as
(-x^2- x) +-(x*sqrt of x+1)

Answer 33

yes

Answer 34

how about teh dennominator...can it stay as (x^2 +2x+1) - (x+1) or that cud be simplifeid?

Answer 35

into (x^2 +x) ?

Answer 36

wat shud i do next?

Answer 37

keep the denominator simplified here try breaking it into this \[\frac{-x^{2}-x - x \times \sqrt{x+1}}{x^{2}+x}\] into \[\frac{-x^{2}-x}{x^{2}+x} + \frac{-x \sqrt{x+1}}{x^{2}+x}\]

Answer 38

now see if you can work with this !

Answer 39

so the integral of teh sum is the sum of its intergrals right?

Answer 40

yes exactly

Answer 41

for teh first part..can u=x^2 adn then du = 2xdx so u can get 1/2d = xdx

Answer 42

im not too sure wat to make u equal to

Answer 43

what if you factored out a -1 from the top of the 1st part? What do you get?

Answer 44

then u get (x^2 +x) / (x^2+x)

Answer 45

thats 1

Answer 46

multiply by what you factored out right? Can you do that integral?

Answer 47

wat do u mean

Answer 48

-1 * 1

Answer 49

\[\int (-1 \times 1) dx\] for the first integral now how do you do the second?

Answer 50

so u factor out a 1- again right to get
\[-\int\limits_{}^{}x \sqrt{x+1} / (x^2+x)\]

Answer 51

now what?

Answer 52

i dont c how anythg wud cancel or simplify

Answer 53

look hard lots of stuff does!

Answer 54

oh so wen u factor (x^2+1) taht becomes x(x+1) which is teh same thing under teh radical

Answer 55

so u can be (x+1)

Answer 56

just do one thing at a time and write out your steps draw a picture if you have to and post it!