Answer
$$\frac{x\ln x}{2} \sqrt{1+(x\ln x)^{2}}+\frac{1}{2} \ln (x\ln x+\sqrt{1+(x\ln x)^{2}})+C$$
Work Step by Step
Given $$\int(\ln x+1) \sqrt{(x \ln x)^{2}+1} d x$$
Let $$u=x\ln x \ \ \ \ \ \ \ \ \ \ du=\ln x+1 $$
Then
$$\int(\ln x+1) \sqrt{(x \ln x)^{2}+1} d x=\int \sqrt{u^{2}+1} d u $$
Use
$$ \int \sqrt{a^{2}+u^{2}} d u=\frac{u}{2} \sqrt{a^{2}+u^{2}}+\frac{a^{2}}{2} \ln (u+\sqrt{a^{2}+u^{2}})+C$$
Then
\begin{align*}
\int(\ln x+1) \sqrt{(x \ln x)^{2}+1} d x&=\int \sqrt{u^{2}+1} d u \\
&=\frac{u}{2} \sqrt{1+u^{2}}+\frac{1}{2} \ln (u+\sqrt{1+u^{2}})+C\\
&=\frac{x\ln x}{2} \sqrt{1+(x\ln x)^{2}}+\frac{1}{2} \ln (x\ln x+\sqrt{1+(x\ln x)^{2}})+C
\end{align*}
