What is the antiderivative of the square root of the quantity (1+x^2)

http://torte.cs.berkeley.edu:8010/tilu

This lookup is very good.

To find the antiderivative of the square root of the quantity (1+x^2), we can use the substitution method.

Let's make the substitution u = 1 + x^2. Then, differentiating both sides with respect to x, we get du/dx = 2x.

Rearranging, we have dx = du/(2x).

Now, substitute u and dx into the original integral:

∫ √(1+x^2) dx = ∫ √u * (du/(2x))

Since dx = du/(2x),

= (1/2) ∫ (√u/u) du

Now, we can simplify the integral by splitting the fraction inside:

= (1/2) ∫ u^(-1/2) du

Integrating u^(-1/2) with respect to u gives us 2 * u^(1/2):

= (1/2) * 2 * u^(1/2) + C

= u^(1/2) + C

Replacing u with the original substitution, we have the final answer:

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

So, the antiderivative of the square root of the quantity (1+x^2) is (1 + x^2)^(1/2) + C.

If you have any further questions or need additional resources, feel free to ask.