$\int \frac{x{\tan }^{-1}x}{\sqrt{1+x^2}}dx=$

$\int \frac{x{\tan }^{-1}x}{\sqrt{1+x^2}}dx$

$=\int \frac{\left(\tan t\right)t}{\sqrt{1+{\tan }^{2}t}}\left({\sec }^{2}tdt\right)$

$\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Put{\tan }^{-1}x=t\Rightarrow x=\tan t\Rightarrow dx={\sec }^{2}tdt$

$=\int \frac{t.\tan t}{\sec t}.{\sec }^{2}tdt$

$=\int t(\sec t.\tan t)dt$

$=t\left(\sec t\right)-\int 1.\sec tdt$

$=t \sec t-\log |\sec t+\tan t|+c$

$=\left({\tan }^{-1}x\right)\sqrt{1+x^2}-\log |\sqrt{1+x^2}+x|+c$