integral of 1/((x^2+1)^2) dx

To find the integral of 1/((x^2+1)^2), you can use a technique called substitution.

Let's start by denoting u = x^2 + 1, then differentiate both sides with respect to x to get du/dx = 2x.

Next, solve for dx:
dx = du/(2x).

Now substitute the values of u and dx into the integral:
∫(1/((x^2+1)^2)) dx = ∫(1/(u^2)) * (1/(2x)) du.

Simplify the expression:
∫(1/(u^2)) * (1/(2x)) du = (1/2) * ∫(1/(u^2x)) du.

Now, we can integrate with respect to u:
∫(1/(u^2x)) du = (1/2) * ∫(u^(-2) * x^(-1)) du.

Integrate (u^(-2) * x^(-1)) with respect to u:
∫(u^(-2) * x^(-1)) du = (1/2) * ∫(1/u^2) du = (1/2) * (-1/u) = -1/(2u).

Finally, substitute u back in terms of x:
-1/(2u) = -1/(2(x^2 + 1)).

Therefore, the integral of 1/((x^2+1)^2) dx is -1/(2(x^2 + 1)) plus a constant of integration.