Example 2.7. $\int_{-1}^{1} \frac{d}{d x}\left(\frac{1}{1+3^{1 / x}}\right) d x$.


Solution. Formal application of the Newton-Leibniz formula is impossible here, since $F(x)=\frac{1}{\left(1+3^{1 / x}\right)}$ is an antiderivative of the integrand everywhere except $x=0$, but this point belongs to the interval of integration $[-1,1]$. To compute this integral, we will first use the additivity property and split the original integral into two integrals over the intervals $[-1,0]$ and $[0,1]$, and then for each indefinite integral, we will write down the corresponding antiderivatives. We have:

$$
\begin{aligned}
& \int_{-1}^{1} \frac{d}{d x}\left(\frac{1}{1+3^{1 / x}}\right) d x= \\
& \quad=\int_{-1}^{0} \frac{d}{d x}\left(\frac{1}{1+3^{1 / x}}\right) d x+\int_{0}^{1} \frac{d}{d x}\left(\frac{1}{1+3^{1 / x}}\right) d x
\end{aligned}
$$

For the first integral, let the antiderivative be $F_{1}(x)$, $x \in[-1,0]$, and for the second integral, let it be $-F_{2}(x), x \in[0,1]$:

$$
\begin{gathered}
F_{1}(x)= \begin{cases}\frac{1}{1+3^{1 / x}}, & \text { for } x \in[-1,0) \\
1 & \text { for } x=0 .\end{cases} \\
F_{2}(x)= \begin{cases}\frac{1}{1+3^{1 / x}} & , \text { for } x \in(0,1] \\
0 & , \text { for } x=0 .\end{cases}
\end{gathered}
$$

Then $I=F_{1}(0)-F_{1}(-1)+F_{2}(1)-F_{2}(0)=1-\frac{3}{4}+\frac{1}{4}-0=\frac{1}{2}$.

In the following examples, possible errors in applying the method of integration by substitution are discussed, and correct solutions are demonstrated.