# Math Help - integral

1. ## integral

Hi,

Could someone please tell me if this is correct so far?

int. x/sq. 1-x^4 dx

I tried u substitution and that didn't work

so, I tried doing integration by parts

u=1-x^4
dv=4x^3

Is the right so far?

Thank you

2. Let $u=x^2$ try that.

3. I put u=x^2+1
dv=x^3
du=2x
v=x^4/4

Would this work? I would then need to it again, right? Also, couldn't I just use this formula? int. du/sq. a^2+u^2=ln(u+sq. a^2+u^2)+c

Thank you

4. Originally Posted by chocolatelover
I put u=x^2+1
dv=x^3
du=2x
v=x^4/4

Would this work? I would then need to it again, right? Also, couldn't I just use this formula? int. du/sq. a^2+u^2=ln(u+sq. a^2+u^2)+c

Thank you
TPH was talking about integration by substitution, not by parts. why are you so eager to jump into integration by parts all the time? it is one of the more tedious integration techniques, you should rather be looking for ways not to use it, and then use it if there are no other options. besides, if you had to use by parts, we would tell you to set u = something and dv = something, here TPH only gave you a u-substitution.

secondly, he said to put u = x^2, not u = x^2 + 1. x^2 + 1 isn't even a part of the problem. follow instructions and see where you get

5. Originally Posted by chocolatelover
int. x/sq. 1-x^4 dx
Before to think on integration by parts, you gotta see first if you can set a substitution.

$\int\frac x{\sqrt{1-x^4}}\,dx$

Now we define a change of variables according to $u=\sqrt{1-x^4}\implies du=-\frac{2x^3}{\sqrt{1-x^4}}\,dx$, the integral becomes to

$\int\frac x{\sqrt{1-x^4}}\,dx=-\frac12\int\frac1{\sqrt{1-u^2}}\,du$

And this is the arcsin function.