how do you integrate x/ (1+x^4) using substitution...???

Answer 1

Let the nastiest bit (1+x^4) = y.
Now, dy/dx = 4x^3. So the integral becomes:
integral of x/y*4x^3 dy.
Note that I haven't changed some xs into ys yet, because something often cancels out.
That is integral of 4x^4/y dy, which is great, because x^4 is just y-1.
So we get integral of 4(y-1)/y dy.
Thats integral of (4y-4)/y dy, ie integral of 4 - 4/y dy. Which becomes 4y - 4 ln|y| + c.
Sub x back in: 4(1+x^4) - 4ln(1+x^4) + c.

Answer 2

I = int [x/(1+x^4)]dx

substitution y = x^2
dy = 2xdx

I = (1/2)int [1/(1+y^2)]dy = (1/2)arctan(y)
= (1/2)arctan(x^2) + C

Answer 3

Let x^2=tg(t), then 2x*dx=dt/(cost)^2, and x*dx/(1+x^4) = 0.5*dt/{(cost)^2*(1+(tgt)^2)} =
0.5*dt/1 = dt/2; I=(1/2)*t = (1/2)*atan(x^2)+C – compare Stephen & me

Answer 4

positioned x = tan ? , dx = sec ² ? ?x^3/(a million+x^2) sx = ? tan ³ ? d? = ? tan ³ ? d? = ? tan ? sec ³ ? d? - ? tan ? d? =a million/2 tan ² ? - ln (sec ?) = a million/2 x ² - ln (a million + tan ² ? )½ = a million/2 x ² -a million/2 ln (a million + x ² )

Answer 5

do your own homework