Find the function f given that the slope of the tangent line

Question

Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point.

f '(x) = 1 − 2x/x² + 1  (0, 6)

Answer 1

I'm going to assume that you meant to use parenthesis around portions of your derivative function and that it is supposed to be:

That means to find the function f(x) that we must integrate:

Let's first break this fraction (and the associated integral) into two:

f(x) = ∫1/(x²+1)dx - ∫(2x)/(x²+1)dx

To make it easier to explain, let's call the problem f(x) = A - B where A = ∫1/(x²+1)dx and B = ∫(2x)/(x²+1)dx

Integrating A is easy if you remember that the derivative of arctan(x) = 1/(x² + 1) which means that the integral of 1/(x² + 1)dx is arctan(x).

To Integrate B, let u = x² + 1, the du/dx = 2x so du = 2xdx meaning that we can write the integral as:

∫1/u du and the integral of this is ln(u). Back-substituting, we get ln(x² + 1).

So f(x) = A - B becomes:

f(x) = arctan(x) - ln(x² + 1) + C

To find the value of C, we can plug in the point (0, 6) so:

6 = arctan(0) - ln(0² + 1) + C

6 = 0 - 0 + C

C = 6

f(x) = arctan(x) - ln(x² + 1) + 6