For example, $\frac{d}{dx} \,\arctan x = \frac{1}{1+x^2}$ and I know this because it is given to me in table of standard derivatives and integrals. But if I want to differentiate something like $\arctan \left(\frac{1-x}{1+x}\right)$, is this the answer?

$$\frac{d}{dx} \,\arctan \left(\frac{1-x}{1+x}\right) = \frac{1}{1+\left(\frac{1-x}{1+x}\right)^2}\cdot \frac{d}{dx}\left( \frac{1-x}{1+x}\right)$$

Can I even do this differentiation at all using the standard derivative given to me? And what about for integration? For example, it is given to me that

$$ \int \tan x \,dx = \ln|\sec x| + C$$

However, is the following correct?

$$ \int \tan (2x-3) \,dx = \frac{1}{2}\,\ln\big|\sec (2x-3)\big| + C $$

I would be grateful if someone could explain the above examples and generalise this topic too.

  • $\begingroup$ When integrating, it's call u-substitution. $\endgroup$ Jan 22, 2015 at 19:39
  • $\begingroup$ Please improve your question by using $\LaTeX$. $\endgroup$ Jan 22, 2015 at 19:39

1 Answer 1


You computation of the differentiation of $arctan(\frac{1-x}{1+x})$ is correct. When differentiating $f(g(x))$ you get $f^{'}(g(x)) g^{'}(x)$. The integral is \begin{equation} \int{tan(2x-3)}=-\frac{1}{2} log(cos(3-2x)) \end{equation}


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