12.5. Let $I_{n}=\int_{\frac{1}{n}}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}, n \geq 2$. Calculate $\lim _{n \rightarrow \infty} I_{n}$.
## Solution.

$$
I_{n}=\int_{\frac{1}{n}}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}=\int_{\frac{1}{n}}^{1} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}+\int_{1}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}
$$

Consider

$$
\begin{aligned}
& \int_{\frac{1}{n}}^{1} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}=\left|\begin{array}{l}
t=\frac{1}{x} \Rightarrow x=\frac{1}{t} \\
d x=-\frac{1}{t^{2}} d t \\
x=\frac{1}{n} \Rightarrow t=n \\
x=1 \Rightarrow t=1
\end{array}\right|=\int_{1}^{n} \frac{d t}{\left.t(t+1)\left(\ln ^{2 n} t+1\right)\right)}= \\
& \int_{1}^{n}\left(\frac{1}{t}-\frac{1}{t+1}\right) \frac{d t}{\ln ^{2 n} t+1}=\int_{1}^{n} \frac{d t}{t\left(\ln ^{2 n} t+1\right)}-\int_{1}^{n} \frac{d t}{(t+1)\left(\ln ^{2 n} t+1\right)}
\end{aligned}
$$

Then

$$
I_{n}=\int_{1}^{n} \frac{d t}{t\left(\ln ^{2 n} t+1\right)}-\int_{1}^{n} \frac{d t}{(t+1)\left(\ln ^{2 n} t+1\right)}+\int_{1}^{n} \frac{d x}{(x+1)\left(\ln ^{2 n} x+1\right)}=\int_{1}^{n} \frac{d t}{t\left(\ln ^{2 n} t+1\right)}
$$

Performing the variable substitution in the last integral, we get

$$
\begin{gathered}
I_{n}=\int_{1}^{n} \frac{d t}{t\left(\ln ^{2 n} t+1\right)}=\left|\begin{array}{c}
z=\ln t \\
d z=\frac{1}{t} d t \\
t=1 \Rightarrow z=0 \\
t=n \Rightarrow z=\ln n
\end{array}\right|=\int_{0}^{\ln n} \frac{d z}{z^{2 n}+1}=\int_{0}^{1} \frac{d z}{z^{2 n}+1}+\int_{1}^{\ln n} \frac{d z}{z^{2 n}+1}= \\
=1-\int_{0}^{1} \frac{z^{2 n} d z}{z^{2 n}+1}+\int_{1}^{\ln n} \frac{d z}{z^{2 n}+1}
\end{gathered}
$$

Estimate each integral separately

$$
\begin{gathered}
0 \leq \int_{1}^{\ln n} \frac{d z}{z^{2 n}+1}<\int_{1}^{\ln n} z^{-2 n} d z=\left.\frac{1}{1-2 n} z^{1-2 n}\right|_{1} ^{\ln n}=\frac{1}{(1-2 n)\left(\ln ^{2 n-1} n\right)}+\frac{1}{2 n-1} \rightarrow 0, n \rightarrow \infty \\
0 \leq \int_{0}^{1} \frac{z^{2 n} d z}{z^{2 n}+1} \leq \int_{0}^{1} \frac{z^{2 n-1} d z}{z^{2 n}+1}=\left.\frac{1}{2 n} \ln \left(z^{2 n}+1\right)\right|_{0} ^{1}=\frac{1}{2 n} \ln 2 \rightarrow 0, n \rightarrow \infty
\end{gathered}
$$

and we obtain: $\lim _{n \rightarrow \infty} I_{n}=1$.