Expert: Scott A Wilson - 1/7/2010

Question
Hello.

What is the antiderivative of the function f(x) = (2+x^2)/(1+x^2)?

Thanks.  

Answer
Here are a few things that need to be known which are discussed below.
1) The antiderivative is the same as the integral, and that is ∫ dx.
2) Integrals have a dx if x is the variable in question.  If it is not x, replace the x with
  whatever it is.  Problems occure with a dy, du, dz, dv, etc., depending if the variable is
  y, u, z, v, etc., respectively.
3) When changing from dx to dy, a conversin factor needs to be found.

The antiderivative is known as the integral.  They are inverse functions of each other.
The derivative of x^n is nx^(n-1), the integral of x^m is x^(m+1)/(m+1).
That is one of the simplest integrals.

To integrate that is not that simple of a problem.




Back to the problem.

First, take f(x) = (1 + 1+x²)/(1+x²), then split it into two fractions.
That is, f(x) = 1/(1+x²) + (1+x²)/(1+x²).

It can be seen that this is the same as f(x) = 1/(1+x²) + 1.

1st Part
------------------------------
The 1st part involves something I didn't see until the 3rd term of calculus,
and that is it uses a right triangle with angle A.  
The far side is x, the near side is 1, and the hypoteneuse is √(1+x²).

When integrating a function, what it is being integrated with respect to is placed at the end.
For example, the integral of f(x) is just ∫f(x)dx.  The 'dx' means we are integrating with
respect to x.  Thus, if we have tan(A) = x, we can take the derivative of both sides.
The derivativeof g(A) = tan(A) is expressed as g'(A) = sec²(A).  The derivative of x is 1.

Since the variables are different, this is a little bit like the chain rule.
At the end goes a dA or dx.  Thus, when differentiating tan(A) = x, we get sec²(A) dA = dx.

It can be seen from the triangle that 1/√(1+x²) is cos(A), so 1/(1+x²) is cos²(A).

The ∫1/(1+x²)dx turns into ∫cos²(A)sec²(A)dA.  It is known that cos(A) = 1/sec(A),
so they cancel and we're left with ∫1 dA, which is just A + K, for some constant K.

Since tan(A) = x, we need to take the inverse tan() of both sides.  This gives A = arctan(x).
Making that substitution of A in terms of x turns A + K gives arctan(x) + K.

The 2nd Part
------------------------------------
The integral of 1 is x + K.


Thus, the entire expression integrates to arctan(x) + x + K.

Explanation of Integrals and K
---------------------------------------------------------------------------
Here, K is just some constant.  It is never used, it is never found, and it is usually ignored,
but it is there just because it is needed.  See, if the function had a +K in it, the derivative of K is 0, so it disappears.  Even when K appears in two different integrals, the sum of thses two different K's is still K.  See, when integrals are evaluated, they are almost always evaluated between 2 points, and the inegral at the lower value is subtracted from the integral at the higher value, thus the answer, as far as the K goes, has a K - K in it.  In this way, it disappears.

If there are any questions you have on integrals (that is antiderivatives), feel free to ask.

Also, one final note.  Derivatives find the slope of a function, integrals find the area beneath the curve between a lower and a upper value.



All this was included since antiderivative is a term used only in calculus for a month or so.
After that, it is known as the integral.