# Evaluate the indefinite integral (x^5)(9+x^6)^1/2 dx?

Evaluate the indefinite integral:

(x^5) [(9+x^6)^1/2] dx

### 4 Answers

- sr_engrLv 610 years agoBest Answer
You could use the wolfram integrator.

(1/9)*(9+x^6)^(3/2) +c

Source(s): [1] 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
Take y=x^6

Then dy = 6*x^5 dx

So your integral becomes:

sqrt(9+y) dy/6

Now take z=9+y

Then dz = dy, so:

sqrt(z) dz/6

This integrates to

[z^(3/2) / 9]

Then replace back your substitutions:

(9+x^6)^(3/2) / 9