∫128▒〖x^4/(1+x^2 ) ⅆx〗

We can rewrite the integral as:

∫128▒〖x^4/(1-x^2+x^2) dx〗

Now we can use partial fractions to separate the integrand:

x^4 / (1-x^2+x^2) = x^4 / (1-x^2+x^2) = x^4 / (1-x^2) + x^4 / x^2

= x^2 + x^2 / (1-x^2)

So, the integral becomes:

∫128▒〖(x^2 + x^2 / (1-x^2)) dx〗

We can now integrate each term separately:

∫128▒〖x^2 dx + ∫128▒〖x^2/(1-x^2) dx〗

The first term is easy to integrate:

∫128▒〖x^2 dx = (1/3)x^3 + C〗

For the second term, we can use u-substitution:

Let u = 1-x^2, then du/dx = -2x and dx = -du/(2x)

So, the integral becomes:

-1/2 ∫128▒〖x^2/(x^2-1) dx〗 = -1/2 ∫128▒〖(u-1)/(2u) du〗

= -1/4 ∫128▒〖(1/u - 1/2) du〗

= -1/4 ln|1-x^2| + 1/8 ln|1-x^2|^2 + C

Putting it all together, we get:

∫128▒〖x^4/(1-x^2) dx = (1/3)x^3 - 1/4 ln|1-x^2| + 1/8 ln|1-x^2|^2 + C〗