Can some one help me integrate
(x^2 / (x^2 + x^4)) dx?
Factor first:
x^2+x^4= x^2(x^2+1)
That will reduce with the numberator to
INT (x^2+1)^-1 dx
To integrate the expression (x^2 / (x^2 + x^4)), you can start by factoring the denominator.
The denominator, x^2 + x^4, can be factored as x^2(x^2 + 1).
Now, you can rewrite the original expression as:
(x^2 / (x^2 + x^4)) = (x^2 / (x^2(x^2 + 1))) = 1 / (x^2 + 1)
Now, you have the integral of (x^2+1)^-1 dx.
To solve this integral, you can use the substitution method. Let u = x^2 + 1, and calculate du/dx = 2x.
Rearrange the equation to solve for dx: dx = du / (2x)
Substitute u and dx into the integral:
∫(x^2+1)^-1dx = ∫(1/u) (du / (2x))
Notice that x appears in the denominator, so you can substitute x = (u-1)^(1/2):
∫(1/u) (du / (2(u-1)^(1/2)))
Now, simplify the integral:
(1/2) ∫ (1/u) (1/(u-1)^(1/2)) du
This integral can be solved by using the substitution u - 1 = t^2:
u = t^2 + 1
du = 2tdt
Substitute u and du into the integral:
(1/2) ∫ (1/(t^2 + 1)) (1/t) 2tdt
= ∫ (1/(t^2 + 1)) dt
This integral is a well-known integral: the arc tangent integral.
The result is:
∫ (1/(t^2 + 1)) dt = arctan(t) + C
Replace t with (u - 1)^(1/2):
arctan((u - 1)^(1/2)) + C
Substituting the original expression for u:
arctan((x^2 + 1 - 1)^(1/2)) + C
= arctan(x) + C
Therefore, the integral of (x^2 / (x^2 + x^4)) is arctan(x) + C, where C is the constant of integration.