Evaluate the indefinite integral (x^5)(9+x^6)^1/2 dx?
Evaluate the indefinite integral:
(x^5) [(9+x^6)^1/2] dx
- sr_engrLv 610 years agoBest Answer
You could use the wolfram integrator.
(1/9)*(9+x^6)^(3/2) +cSource(s):  http://integrals.wolfram.com/index.jsp?expr=%28x^5...
- Anonymous10 years ago
= (1/9)(9+x^6)^(3/2) + c
- currenLv 43 years ago
hi guy decrease than me, the imperative is dx/(5-3x) not dx/5 - 3x. that is you who's erroneous. Did you even study the question? the indefinite imperative could be (a million/-3)*ln(5-3x) + C or written as ln(5-3x)/-3 + C you are able to verify it by applying computing the by-product of the respond. you need to get the decrease back what you began with. in case you ever have a function like one you asking approximately (the place you have dx divided by applying something with a variable aka "x" to the flexibility of a million). as long simply by fact the "x" is to the flexibility of a million, then you will comprehend that the anti-by-product is going to be some style of ln() function. ex: given the function (a million/x)dx or dx/x, the antiderivative could merely be ln(x).
- 10 years ago
Then dy = 6*x^5 dx
So your integral becomes:
Now take z=9+y
Then dz = dy, so:
This integrates to
[z^(3/2) / 9]
Then replace back your substitutions:
(9+x^6)^(3/2) / 9