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

[brutalmadness]

Homework Statement

\intx(3x^2+5)^1/4

2. The attempt at a solution
u=3x^2+5 du=6x
1/6\int6x(3x^2+5)^1/4

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

1/6\intdu(u)^1/4
1/6\int4/5(u)^5/4
\int2/15(3x^2+5)^5/4

and then i get stuck.

 

[Alienjoey]

Homework Statement

\intx(3x^2+5)^1/4

2. The attempt at a solution
u=3x^2+5 du=6x
1/6\int6x(3x^2+5)^1/4

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

1/6\intdu(u)^1/4

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)

1/6\int4/5(u)^5/4
\int2/15(3x^2+5)^5/4

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

 

[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).

 

[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\intu^(1/4)du

Rather, another way for me to put it:

is \intudu = \int0.5u^2du ?

 

[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...

 

[Alienjoey]

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

1/6\intdu(u)^1/4 = 1/6\int4/5(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.

 

[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.