Find the anti derivative for 4/(x^2+1)
4 (x^2 + 1)^-1
4 * -1 * (x^2 + 1)^-2 * 2x
-8x / (x^2+1)^2
Your question is of this type:
Integral of du / (a^2 + u^2) = (1/a) tan-1(u/a) +c
OOPS... missed the "anti" part
that is the derivative
To find the antiderivative of the function 4/(x^2+1), we can use the technique of integration. In this case, we can use the substitution method.
Let's start by letting u = x^2 + 1. Then, we can find du/dx by taking the derivative of both sides with respect to x:
du/dx = d/dx (x^2 + 1)
du/dx = 2x
Now, we can rewrite the function in terms of u:
4/(x^2 + 1) = 4/u
Next, we can rewrite du/dx in terms of du:
du = 2x dx
To solve for dx, we can divide both sides by 2x:
dx = du/(2x)
Substituting these expressions into the integral, we have:
∫ (4/(x^2+1)) dx = ∫ (4/u) (du/(2x))
Now, we can simplify the integral:
∫ (4/u) (du/(2x)) = (1/2) ∫ (4/u) (du/x)
Notice that 4 is a constant and can be factored out of the integral:
(1/2) ∫ (4/u) (du/x) = 2 ∫ (1/u) (du/x)
Now, we can divide both sides of the integral by x:
2 ∫ (1/u) (du/x) = 2 ∫ (1/u) (du/u)
Integrating (1/u) with respect to u gives us:
2 ∫ (1/u) (du/u) = 2 ln|u| + C
Since we initially let u = x^2 + 1, we can substitute back in:
2 ln|u| + C = 2 ln|x^2 + 1| + C
Therefore, the antiderivative of 4/(x^2+1) is 2 ln|x^2 + 1| + C, where C is the constant of integration.