For determining the antiderivative $F(x)$ of a given real function $f(x)$ in a ``closed form'', i.e. for integrating $f(x)$ , the result is often obtained by using the

Theorem 1   If $$\int f(x)\,dx = F(x)+C$$ and $x = x(t)$ is a differentiable function, then

(1)

Proof. By virtue of the chain rule, $$\frac{d}{dt}F(x(t)) = F'(x(t))\cdot x'(t),$$ and according to the supposition, $F'(x) = f(x)$ . Thus we get the claimed equation (1).

Remarks.

• The expression $x'(t)\,dt$ in (1) may be understood as the differential of $x(t)$ .
• For returning to the original variable $x$ , the inverse function $t = t(x)$ of $x(t)$ must be substituted to $F(x(t))$ .
  

Example. For integrating $\int \frac{x\,dx}{1+x^4}$ we take $x^2 = t$ as a new variable. Then, $2x\,dx = dt$ , $x\,dx = \frac{dt}{2}$ , and we get $$\int \frac{x\,dx}{1+x^4} = \frac{1}{2}\int \frac{dt}{1+t^2} = \frac{1}{2}\arctan t+ C= \frac{1}{2}\arctan x^2+C.$$