How to prove that Integral S 1/(a^2+x^2) dx= 1/a arctan(x/a) + C ?

In order to answer this question, at first we need to use the method of substitution. That means, we're trying to replace x with a different variable u. Lets use the substitution of x = a tan u, then du/dx = a (sec u)^2. From that we can substitute both x and dx with the new variable u. As such S 1/(a^2 +x^2) dx becomes S 1/(a^2 + (a tan u)^2 * a (sec u)^2 du, or S a (sec u)^2/ a^2 (1 +(tan u)^2 ) du. Using tigonometric identities, we can simplify 1+(tan u)^2 to (sec u)^2 obtaining S (1/a) * (sec u)^2/ (sec u)^2 du = 1/a S 1 du. That would be equal to 1/a * u +c.Last part of the question is how to find u. Since we know that x = a tan u, we also know that x/a = tan u, that means that for arctan (tan u) = arctan (x/a) and thus, u = arctan (x/a). Therefore the S 1/(a^2 +x^2) dx = 1/a * u +c =(1/a) arctan (x/a) +C