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# The integration of $\int {x.\dfrac{{\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}} dx$ is equal to:   $\left( A \right)$ $\sqrt {1 + {x^2}} \ln \left( {x + \sqrt {1 + {x^2}} } \right) - x + c$ $\left( B \right)$ $\dfrac{x}{2}.{\ln ^2}\left( {x + \sqrt {1 + {x^2}} } \right) - \dfrac{x}{{\sqrt {1 + {x^2}} }} + c$    $\left( C \right)$ $\dfrac{x}{2}.{\ln ^2}\left( {x + \sqrt {1 + {x^2}} } \right) + \dfrac{x}{{\sqrt {1 + {x^2}} }} + c$ $\left( D \right)$ $\sqrt {1 + {x^2}} \ln \left( {x + \sqrt {1 + {x^2}} } \right) + x + c$  Hint: In these types of functions we will try our best to make the integral in simplest form and this can be done by method i.e. rationalization and substitution. Generally we will do substitution so that one function would become derivative of another, so both can be substituted at the same time.

Complete step by step solution:
Given, $\int {x.\dfrac{{\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}} dx$
Now, we will substitute $x$with $\tan \theta$
So, let $x = \tan \theta$
So we get,
$dx = {\sec ^2}\theta d\theta$
And substitute this to the equation we get
$= \int {\tan \theta .\dfrac{{\ln \left( {\tan \theta + \sqrt {1 + {{\tan }^2}\theta } } \right)}}{{\sqrt {1 + {{\tan }^2}\theta } }}} \times {\sec ^2}\theta d\theta$
We can know that, $1 + {\tan ^2}\theta = {\sec ^2}\theta$
$\sqrt {1 + {{\tan }^2}\theta } = \sqrt {{{\sec }^2}\theta } \\ \sqrt {1 + {{\tan }^2}\theta } = \sec \theta \\$
So when we put it to the equation we get,
$= = \int {\tan \theta .\dfrac{{\ln \left( {\tan \theta + \sec \theta } \right)}}{{\sec \theta }}} \times {\sec ^2}\theta d\theta$
$= \int {\tan \theta .\ln \left( {\tan \theta + \sec \theta } \right)} .\sec \theta d\theta$
$= \int {\sec \theta \times \tan \theta \times \ln \left( {\tan \theta + \sec \theta } \right)} d\theta$
Now, we will apply Integration by parts
Our first function $= \ln \left( {\tan \theta + \sec \theta } \right)$
Second function $= \sec \theta .\tan \theta$
We know $\int {\sec \theta .\tan \theta } d\theta = \sec \theta$
Now we put it to the equation, $\ln \left( {\tan \theta + \sec \theta } \right) \times \int {\sec \theta .\tan \theta d} \theta - \int {\dfrac{{d\left( {\ln \left( {\tan \theta + \sec \theta } \right)} \right)}}{{d\theta }}\int {\sec \theta .\tan \theta d\theta } }$
$=$$\ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta - \int {\dfrac{1}{{\tan \theta + \sec \theta }} \times \left( {{{\sec }^2}\theta + \sec \theta \tan \theta } \right)} \times \sec \theta d\theta$
$= \ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta - \int {\dfrac{1}{{\left( {\tan \theta + \sec \theta } \right)}} \times \sec \theta \left( {\sec \theta + \tan \theta } \right)} \times \sec \theta d\theta$
$= \ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta - \int {{{\sec }^2}} \theta d\theta$
$= \ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta - \tan \theta + c$
We get our answer only when we replace $\theta$in terms of $x$
Now replace $\theta$ in term of $x$
$\sqrt {1 + {x^2}} \ln \left( {x + \sqrt {1 + {x^2}} } \right) - x + c$
So, the correct option is $A$.

Note: Generally one can make mistakes during the substitution of $dx,dy$etc. So be careful during substitution. Always converts the final answer after substitution into the same variable as it was given in the question.
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