# Thread: another difficult integral

1. ## another difficult integral

Hi everyone,

Could someone please tell me how I would solve the integral 1 to infinity x/sq.1+x^6?

I did separation by parts?

u=sq. 1+x^6
dv=xdx
du=1/2(1+x^6)(6x^5)dx
v=x^2/2

Would this work?

Thank you very much

2. that would work, you would get

$\dfrac{x^2}{\sqrt{1+x^6}}-\dfrac{3}{7}\int x^7\,dx$

3. Careful there, integration by parts is not well applied.

4. sry...man i did wrong i was reading your question n doing my own at the same time....

5. Originally Posted by chocolatelover
Hi everyone,

Could someone please tell me how I would solve the integral 1 to infinity x/sq.1+x^6?

I did separation by parts?

u=sq. 1+x^6
dv=xdx
du=6x^5ddx
v=x^2/2

Would this work?

Thank you very much
$\frac d{dx} \sqrt{1 + x^6}$ is NOT $6x^5$

6. I got lim as t goes to infinity [sq. 1+x^6(x^2)/(2)-1/4int. 6x^12+6x^7]1 to t

is that correct?

Thanks

7. Then I got lim as t goes to infinity [sq. 1+t^6(t^2/2)-1/4((6t^13/13+6(t)^8/8]-[sq.1+1^6(1^2)/(2)-1/4(6(1)^13/13+6(1)^8/8]

Is that correct?

I wouldn't have use L'Hopital's rule, would I? It would just be infinity, right?

Thank you

8. Just for your benefit, I don't believe this is integrable by elementary means.
Therefore, parts will not work too well.

$\int_{1}^{\infty}\frac{x}{\sqrt{1+x^{6}}}dx$

I ran it through Maple and got 0.9473799840

For the indefinite, I got: $\frac{x^{2}}{2}hypergeom\left([\frac{1}{3},\frac{1}{2}],[\frac{4}{3}],-x^{6}\right)$

Just a thought to throw out there.

9. did you check to see if there is actually an answer because i dont think this function can even be integrated....

10. Originally Posted by galactus
Just for your benefit, I don't believe this is integrable by elementary means.
Therefore, parts will not work too well.

$\int_{1}^{\infty}\frac{x}{\sqrt{1+x^{6}}}dx$

I ran it through Maple and got 0.9473799840

For the indefinite, I got: $\frac{x^{2}}{2}hypergeom\left([\frac{1}{3},\frac{1}{2}],[\frac{4}{3}],-x^{6}\right)$

Just a thought to throw out there.
Well, seeing as the indefinite integral has a solution the definite integral ought to as well. The hypergeometric function is a series solution so I wonder if chocolatelover should just derive a Taylor series for the integrand and integrate term by term? Of course then you have to worry about series convergence, etc, so it wouldn't be a walk in the park.

-Dan

11. As I stated previously, the solution is 0.9473799840.

Of course, I arrived at this via Maple 10.

Here's a graph using Simpson's rule and 100 partitions. I couldn't use 1 to infinity, so I just used 1 to 100. The graph gets infinitely close to the x-axis, so it's not that bad of an estimate