How do I find the coordinates of the two points on the curve y=ln(x^2+1) where the gradient is 5/13

To find the coordines of the 2 points on the curve  y =
ln(x^2+1), we  find dy/dx and equate it to 5/13 and solve for x. Then the solution value
of the x will be put in y = ln(x^2+1)to find the y coordinate
value.

dy dx = {ln(x^2+1} = {1/(x^2+1)}( href="mailto:x@+1)'">x^2+1)' = 2x/(x^2+1)

So
 2x/(x^2+1)  = 5/13 and solve for x.

Multply by 
13(x^2+1)

26x =5(x^2+1)

We
write this as a quadratic equation:

5x^2-26x +5=
0.

(5x-1)(x-5) = 0

5x-1 = 0 .
Or x - 5 =0

Or x = 1/5 . Or x=
5.

So to get yhe  corresponding y coordinates we put these 
solution values of x in  y = ln(x^2+1)

When x = 1/5 , y =
ln{(1/5)^2+1) = ln (26/25)

When x = 5, y = ln(5^2+1) =
ln26.

Therefore the slope is 3/13 for the curve y =
ln(x^2+1) at (1/5 , ln(26/25) and  (5 , ln25).