Question

how to find the antideriative of this?

Answer 1

\[\int\limits (x^3+5)^7\]
can you only do it by expanding it first and then finding the antiderivative of the terms one by one?
or there is simpler way?

Answer 2

it should be (x^3 + 5)^7 dx

Answer 3

you are not supposed to expand it i think you add one to the power divide by the power of the bracket so ((x^3+5)^8)/8 and also you must do the same to the x so final answer I think is(((x^4/4)+5x)^8)/8

Answer 4

+c

Answer 5

naah, that doesn't work
and not i don't think there is any other way

Answer 6

I already tried u-substitution. but I think it cannot be applied here because
du will be equal to 3x^2 dx
and there are no 3x^2 on the given
and I think you are not allowed to multiply 3x^2 to compensate for the 3x^2 missing in du
I hope you understand what I'm saying. :)

Answer 7

yes, u substitution or trig subs. will not work
only way is to expand and get 8 terms and then integrate

Answer 8

I will look it up but I think substitution works but need to check which way

Answer 9

oh thanks,
so when the given is something like
(x^n + c)^a
where n, a = integer >1
the only thing you can do is expand it?

Answer 10

if nothing else is multiplied with it, then yes.

Answer 11

ok thanks
what if it is looks like this:\[\int\limits x(x^7+2)^8 dx\]
can you do u-substitution here or something else?

Answer 12

if it were x^6 (x^7+2)^8
only then you can use a u-substitution

Answer 13

you can do it in this case because you have x this means that 7x*6 which is the differentiation of x^7 can be written as 6/7th of 7/6x

Answer 14

i did not get that ^ sorry

Answer 15

I think I agree with hartnn that it cannot be done.
is it because your du will be
du = 7x^6 dx
and you cannot multiply 7x^5 to compensate for the missing term in the given?
am I right?

Answer 16

correct!
and even if you multiply AND divide x^5 to get a x^6 term, you would still have x^5 in the denominator which is unaccounted for.

Answer 17

pull out one of the factors and use it as a polynomial

Answer 18

(x^3 + 5)^7

(x^3 + 5) (x^3 + 5)^6

Answer 19

Thanks
Why is it you cannot multiply a variable in algebra or here?

Answer 20

because variables are not constant

Answer 21

what's the difference?

Answer 22

lol
you can multiply constant as that will not change your function.
you cannot multiply variables because that may introduce a new zero or pole on the function

Answer 23

like x+2 and x(x+2)/x are different functions

Answer 24

the latter cannot have 0 in its domain

Answer 25

ohhhhh, thanks
I understand it now :)

Answer 26

welcome ^_^

Answer 27

@amistre64

What should I do with
(x^3 + 5) (x^3 + 5)^6
I think it makes no difference?

Answer 28

i had a thought about using by parts, but afterwards i seemed a bit futile

you mioght be able to implement a binomial setup:

(x+y)^n = \(\large \sum\binom{n}{k}x^ny^{n-k}\)

Answer 29

but thats just a fancy way of expanding it

Answer 30

ohhh, Thanks for the idea. I think that will be our lesson in the future. We've only talked about the basic integration.
Thanks to all of you who helped.
:)

Answer 31

looks interesting idea
think it will work@amistre62

Answer 32

yeah, but its only a compact way of showing the expansion :/

Answer 33

I know but I thought I remembered another way but have not found it so perhaps that is the best solution

Answer 34

http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx
useful website