find antiderivative

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# find antiderivative

Submitted by nona_nonomo on Tue, 10/26/2010 - 04:17

Forums:

find antiderivative of a function f (x) = x ^ 6 / (1 + x ^ 12)

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## Versions

(v1) by nona_nonomo 2010-10-26

## Re: find antiderivative

i know result but i donot know how to solve this problem step by step

## Re: find antiderivative

i want to know that how is the solving step by step

## Re: find antiderivative

I will at least get you started then.

You need to factor the denominator into quadratic factors and then convert to partial fractions. The rest is pretty much paint by numbers math. To get you started

xÂ¹Â² + 1 = (xÂ² + âˆš2 x + 1)*(xÂ² - âˆš2 x + 1)*

(xÂ² + ((âˆš3 - 1)/âˆš2)x + 1)*(xÂ² - ((âˆš3 - 1)/âˆš2)x + 1)*

((xÂ² + ((âˆš3 + 1)/âˆš2)x + 1)*(xÂ² - ((âˆš3 + 1)/âˆš2)x + 1)

How did I do that? Use

(xÂ² + ax + 1)((xÂ² - ax + 1) = x^4 + (2 - aÂ²)xÂ² + 1)

## Re: find antiderivative

I will at least get you started then.

You need to factor the denominator into quadratic factors and then convert to partial fractions. The rest is pretty much paint by numbers math. To get you started

x^12 + 1 = (x^2 + Ax + 1)*(x^2 - Ax + 1)*

(x^2 + Bx + 1)*(x^2 - Bx + 1)*

(x^2 + Cx + 1)*(x^2 - Cx + 1)

Where A = sqrt(2)

B = (sqrt(ÂˆÂš3) - 1)/sqrt(2)

C = (sqrt(ÂˆÂš3) + 1)/sqrt(2)

How did I do that? Use

(x^2n + ax^n + 1)((x^2n - ax^n + 1) = x^4n + (2 - a^2)x^2n + 1)

## Re: find antiderivative

thanks a lot. let me consider.

## Re: find antiderivative

thank you very much! i am afraid that this function under cannot regard that (u.du)/(1+u^2) because dx is not equal to du (du=6x^5.dx)

Please help me to solve this problem. thank you a lot

## Re: find antiderivative

Hi,

you are obviously right, I did checked the $du$, $dx$ part.

I shall chek it again ........ and see what can be done

Sorry

## Re: find antiderivative

http://integrals.wolfram.com/index.jsp

## Re: find antiderivative

Let $u = x^{6}$ then the function under consideration assumes the form

$\hat{f}(u) = \frac{u}{1+u^{2}}$ and it is easy to get that an antiderivative of $\hat{f}$ is given by the

$\hat{F}(u) = \frac{1}{2}ln (1 + u^2) + c$, $c$ a constant.