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I'm trying to study for a Calc test tomorrow but i'm really having problem understanding one part of this problem. I already came up with the right aswer, but only b/c the multiple choice questions kinda gave it away, I really need help plz

Evaluate the integral:

integral [ x/(81+x^4)] dx
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Here is where the problem starts...i know that i have to let 'x' equal to something, but I dont know how to do that. Once again, the only reason why I was able to get tthe right answer was by looking at the multiple choice answers and figuring it out from there...

x= sqrt(tan 9t) { this implies that t= arctan (x^2)/9 }
dx= (9sec^2 9t)*[1/2 (tan 9t)^(-1/2)] dt

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thus....
integral [ x/ (81+x^4)] dx=
integral [ sqrt(tan 9t)/ (81-(tan 9t)^2 )] * [1/ 2sqrt(tan9t)]* 9sec^2 (9t)

And if simplify you get
(1/18)arctan [(x^2)/9] + C

2007-02-20 19:28:03 · 4 answers · asked by Sir Excalibur 2 in Science & Mathematics Mathematics

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NORTHSTAR if you are still on here, when you do the following...

Integrate ∫[x/(81+x^4)]dx.
Let
9tanθ = x²
9sec²θdθ = 2xdx
(9/2)sec²θdθ = xdx

how do you know to let ' 9tanθ = x² ', this is what im having problems with

2007-02-20 19:53:12 · update #1

4 answers

Integrate ∫[x/(81+x^4)]dx.
Let
9tanθ = x²
9sec²θdθ = 2xdx
(9/2)sec²θdθ = xdx

∫[x/(81+x^4)]dx
= (9/2)∫[sec²θ/(81+81tanθ²)]dθ
= (9/2)∫[sec²θ/(81secθ²)]dθ
= (1/18)∫dθ
= θ/18 + C

9tanθ = x²
tanθ = x²/9
θ = arctan(x²/9)

= θ/18 + C
= (1/18)arctan(x²/9) + C

============================

When I see something of the form
a² + x²

I consider using the following trig identity as one possibility.
1 + tan²θ = sec²θ

In this case we need
81 + 81tan²θ = 81sec²θ

Look at it this way.
9² + (x²)² = 9² + (9tanθ)² = (9secθ)²

2007-02-20 19:37:37 · answer #1 · answered by Northstar 7 · 1 0

put u=x^2 which leads to xdx = 1/2 du

Your integral then reduces to

1/2 Integral(du/(81 + u^2) = 1/18 arctan(u/8) = 1/18arctan(x^2/9)

I have ignored the integration constant - you can't in the test.

2007-02-21 03:39:19 · answer #2 · answered by A S 4 · 0 0

Int[x/81+x^4)]dx

Let 9t be x^2!
then dt=2/9xdx

and
Int[x/81+x^4)]dx=
=Int[9/2(81+81t^2)]dt
=1/18arctantdt=
=1/18arctan(x^2/9)+C

wishing you be happy with this answer

2007-02-21 04:24:21 · answer #3 · answered by happyrabbit 2 · 0 0

its been awile but i would think if youve taken the derivative of the answer
and got x/(81+x^4) then you got it wired
that is the only sure fire way to check your answer as there isnt any function that
the derivative cant be easly found.

2007-02-21 03:40:11 · answer #4 · answered by charliefromwashington 3 · 0 0

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