integration of 6x^2/(1+x^4) w.r.t. x?????????

Answer 1

Like gianlino said, partial fractions....yuk! But it is good to practice some.

Or technology, especially once you understand the concepts. You should end up with (Maple output):

3/4*2^(1/2)*[
ln((x^2-x*2^(1/2)+1)/(x^2+x*2^(1/2)+1))
+2arctan(x*2^(1/2)+1)
+2arctan(x*2^(1/2)-1)]

It looks better in "pretty print" but then that didn't paste into this window well.

I can't even paste it in messy-way in this window. Anyway, computer algebra systems (CAS) will do this automatically.

Answer 2

So you want to find

Integral (6x^2 / (1 + x^4)) dx =

Integral ( [1/(1+x^4)] [6x^2]dx )

This would be one way to solve it. An easier way might exist though. We're going to first use substitution.

Let u = x^3. This means that
du = 3x^2 dx, and
2du = 6x^2 dx

Since u = x^3, then u^(4/3) = x^4, so we substitute appropriately as:

Integral ( [1/(1+u^(4/3))] du )

At this point, I gotta be honest and say I don't know how to solve this one. I thought I'd get somewhere with this and found out I couldn't.

Answer 3

F(x) = 6 (x^2)/(1+ x^4)

put x^2 = t then 2xdx = dt

Substitute this in the original function

f(t) = (6.t) dt /((1+t^2)*2*t^1/2)

Then its a simple integration of products.

Hope that helps!!!!

Answer 4

You have to write (1+x^4) as (1+x^2)^2 - 2x^2 ans write it as a product to reduce the rational fraction and wind up with terms like

( (ax+b) / ( 1 + sq{2}x + x^2) ) + ( (a'x+b') / ( 1 - sq{2}x + x^2) )

where sq denotes square root. You have to identify a, a' b, b' and you wind up with a combination of ln and arctan. OK?

Answer 5

So y=6xx/(1+xxxx), first let’s factorize x^4+1;

Formulas Reminder: j=sqrt(-1); cos(t)+jsin(t)=exp(j*t);
1 = cos(pi) + jsin(pi) = exp(j*pi) or
1 = cos(-pi) + jsin(-pi) = exp(-j*pi); thus
4 complex roots are: x1=-exp(j*pi/4), x2=+exp(j*pi/4),
x3=-exp(-j*pi/4), x4=+exp(-j*pi/4); while
exp(+jpi/4) = cos(pi/4) + j*sin(pi/4) = 1/sqrt(2) + j/sqrt(2) and
exp(-jpi/4) = cos(-pi/4) + j*sin(-pi/4) = 1/sqrt(2) – j/sqrt(2);

hence x^4+1 = (x-x1)*(x-x3) * (x-x2)*(x-x4) or multiplying pairs
back: (1) xx -(x1+x3)x +x1x3 and (2) xx -(x2+x4)x +x2x4;
here (x1+x3) = -1/sqrt2-j/sqrt2 – 1/sqrt2+j/sqrt2 = -sqrt2; x1x3=1;
here (x2+x4) = 1/sqrt2+j/sqrt2 + 1/sqrt2-j/sqrt2 = +sqrt2; x2x4=1;
So y=6xx /(xx-sqrt2x+1) /(xx+sqrt2x+1) =
= (Ax+B) /(xx-sqrt2x+1) + (Cx+D) /(xx+sqrt2x+1), where A B C D are to be found;

Adding both ratios back we get the numerator:
6xx = (Ax+B)(xx+sqrt2x+1) + (Cx+D)(xx-sqrt2x+1), or
6xx = Axxx+Bxx+sqrt2Axx+sqrt2Bx+Ax+B +
+ Cxxx+Dxx-sqrt2Cxx-sqrt2Dx+Cx+D ;

Grouping powers:
(xxx) A+C=0; (xx) B+sqrt2A+D-sqrt2C =6;
(x) sqrt2B+A-sqrt2D+C=0; (x^0) B+D=0;
Then (xx) (B+D)+sqrt2(A-C)=6,
hence A-C=3sqrt2, hence A=3/sqrt2 C=-3/sqrt2;
Then (x) (A+C)+sqrt2(B-D)=0; B=D=0;

Thus I=I1+I2, where
I1 = +(3/sqrt2) * int {x/(xx-sqrt2x+1)} dx;
I2 = -(3/sqrt2) * int {x/(xx+sqrt2x+1)} dx;

Let y1 = x/(xx-sqrt2x+1) and u/sqrt2=x-sqrt2/2,
then dx=du/sqrt2 and y1=sqt2(u+1)/(uu+1);
thus int {(u+1)/(uu+1)} du = 0.5ln(uu+1) + atan(u);

Let y2 = x/(xx+sqrt2x+1) and v/sqrt2=x+sqrt2/2,
then dx=dv/sqrt2 and y2=sqt2(v-1)/(vv+1);
thus int {(v-1)/(vv+1)} dv = 0.5ln(vv+1) - atan(v);

At last
I = (3/sqrt2) * {0.5ln((uu+1)/(vv+1)) + atan((u+v)/(1-uv))};
here (uu+1)/(vv+1) = (xx-sqrt2x+1) / (xx+sqrt2x+1);
and (u+v)/(1-uv) = sqrt2x/(1-xx);
I = (3/sqrt2) * {0.5*ln((xx-sqrt2x+1) / (xx+sqrt2x+1)) + atan(sqrt2x/(1-xx))};

Differentiate back to check me!