Replace all expressions containing the derivative of the arctangent. If the arctan appears as a derivative, as in d/dx(arctan(ax)), you can easily substitute, because the derivative of the arctan is 1/(1 + x^2).
Replace the arctan by its infinite series expansion, discovered by James Gregory in the 17th century.
arctan (x) = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 -....
Determine the required accuracy of the problem, and cut off the infinite series at the point where that accuracy is exceeded. That is, continue to add terms to the series until a term is smaller than your desired accuracy. For example, if 1 percent accuracy is acceptable for a problem in which x = .74, you would keep adding terms to the expansion until a term is smaller than 1 percent of 0.74. The fifth term, x^9/9 equals .00739. Dividing that by .74 results in an answer smaller than .001. Each additional term will be smaller and you can ignore them and reach the desired accuracy of 1 percent.