Can the antiderivate of $\frac{1}{(x^2+1)^2}$ be calculated without using the iteration formula? To calculate the antiderivate of $$\frac{1}{(x^2+1)^2}$$ , we can either use the iteration formula reducing the exercise to the integral of $\ \frac{1}{x^2+1}\ $ or we can use $$(\frac{x}{x^2+1})'=\frac{1-x^2}{(x^2+1)^2}$$ , but I do not see how we can get the antiderivate of $\ \frac{1-x^2}{(x^2+1)^2}\ $ either without any guess (If we know this antiderivate , we can express $\frac{1}{(x^2+1)^2}$ as a linear combination of $\ f(x):=\frac{1}{x^2+1}\ $ and $\ g(x):=\frac{1-x^2}{(x^2+1)^2}\ $ , namely $\ \frac{1}{(x^2+1)^2}=\frac{f(x)+g(x)}{2}\ $)

How can I calculate $\int \frac{dx}{(x^2+1)^2}$ only by using intgration by parts and the substitution rule as well as other basic integration rules ? I am looking for a solution not containing a guess or the iteration formula.


We have
$$\int\frac{dx}{(1+x^2)^2}=\int \frac{1+x^2-x^2}{(1+x^2)^2}dx=$$
$$\arctan(x)+\frac{1}{2}\int x\frac{-2x}{(1+x^2)^2}dx=$$
$$\arctan(x)+\frac{1}{2}\left(\left[x \frac{1}{1+x^2}\right]-\int \frac{1}{1+x^2}\right)=$$
$$\frac{1}{2}\arctan(x)+\frac{x}{ 2(1+x^2)  }+C$$
