How to find the integral of x(3x^2 + 5)^1/4

1. Apr 28, 2008

brutalmadness

1. The problem statement, all variables and given/known data
$$\int$$x(3x^2+5)^1/4

2. The attempt at a solution
u=3x^2+5 du=6x
1/6$$\int$$6x(3x^2+5)^1/4

this is where i'm getting sloppy, i think...

1/6$$\int$$du(u)^1/4
1/6$$\int$$4/5(u)^5/4
$$\int$$2/15(3x^2+5)^5/4

and then i get stuck.

2. Apr 28, 2008

Alienjoey

Good so far...(except du=6xdx and you forgot to include a few dummy variables throughout the problem, but you are accounting for them in your attempt--just remember to write them down too)

Where are you getting the integral sign from in this part of the problem?

3. Apr 28, 2008

brutalmadness

i was trying to use the power rule for integrals (x^n+1/n+1). but i have a feeling i can't do that. haha :)

then i pulled the 1/6 back into it and multiplied it out (1/6 times 4/5=2/15).

4. Apr 28, 2008

Alienjoey

No, you were doing it right, no need to second guess yourself. The power rule for integrals is all you need to solve this.

Try to integrate the following again:

1/6$$\int$$u^(1/4)du

Rather, another way for me to put it:

is $$\int$$udu = $$\int$$0.5u^2du ?

Last edited: Apr 28, 2008
5. Apr 28, 2008

brutalmadness

no, they wouldn't be equal. for example, does 5=0.5(5)^2? no; 5 does not equal 12.5

did i pull my denominator up incorrectly? if (x^n+1/n+1) is (x^.25+1/.25+1), then (x^1.25/1.25). so to pull up the denominator...

6. Apr 28, 2008

Alienjoey

Ok, if those two aren't equal, then why would

1/6$$\int$$du(u)^1/4 = 1/6$$\int$$4/5(u)^5/4

Your denominator and everything is fine, you just have an extra integral sign floating around after you already correctly evaluated the integral. I didn't outright say it since I thought you'd notice it, but I suppose my references were a bit too vague.

Once you get rid of the extra integral sign at the point where you are "stuck", you would have 2/15(3x^2+5)^5/4 (+C), and that would simply be your final answer, unless there was an initial condition or something.

7. Apr 28, 2008

brutalmadness

oh wow, that's what i get for trying to do calculus late at night. i was trying to evaluate an integral after i already evaluated it. so my final answer was correct... just need to watch out and make sure i drop the integral sign.

should have been...
(1/6)(4/5)u^5/4
2/15(3x^2+5)^5/4+C

thank you so much.

Last edited: Apr 28, 2008
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