Question

Given, $\int {x.\dfrac{{\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}} dx$ is equal to
$\left( 1 \right){\text{ }}\sqrt {1 + {x^2}} .\ln \left( {x + \sqrt {1 + {x^2}} } \right) - x + c$
$\left( 2 \right){\text{ }}\dfrac{x}{2}.{\ln ^2}\left( {x + \sqrt {1 + {x^2}} } \right) - \dfrac{x}{{\sqrt {1 + {x^2}} }} + c$
$\left( 3 \right){\text{ }}\dfrac{x}{2}.{\ln ^2}\left( {x + \sqrt {1 + {x^2}} } \right) + \dfrac{x}{{\sqrt {1 + {x^2}} }} + c$
$\left( 4 \right){\text{ }}\sqrt {1 + {x^2}} .{\ln ^2}\left( {x + \sqrt {1 + {x^2}} } \right) + x + c$

Answer 1

Substitute $x = \tan \theta$ and $dx = {\sec ^2}\theta d\theta$ . Then simplify the expression and evaluate the integral by using formulas and values. The $uv$ formula of integration is as $\int {uvdx = u\int {vdx{\text{ }} - {\text{ }}\int {u'\left( {\int {vdx} } \right)} } } dx$ . By using this formula you can solve it further and get the desired result.

Complete step by step answer:
The given function is $\int {x.\dfrac{{\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}} dx$ ------- (i)
Let $x = \tan \theta$ . On differentiating it with respect to $\theta$ we get $\dfrac{{dx}}{{d\theta }} = {\sec ^2}\theta$ . And from this the value of $dx = {\sec ^2}\theta d\theta$
So the equation (i) becomes, $\int {\tan \theta .\dfrac{{\ln \left( {\tan \theta + \sqrt {1 + {{\tan }^2}\theta } } \right)}}{{\sqrt {1 + {{\tan }^2}\theta } }}} .{\sec ^2}\theta d\theta$ ------- (ii)
We know that ${\tan ^2}\theta + 1 = {\sec ^2}\theta$ . Therefore we can write the equation (ii) as
$\int {\tan \theta .\dfrac{{\ln \left( {\tan \theta + \sec \theta } \right)}}{{\sec \theta }}} .{\sec ^2}\theta d\theta$
Now $\sec \theta$ will be cancelled out by $\sec \theta$ and we get
$\int {\tan \theta .} \ln \left( {\tan \theta + \sec \theta } \right).\sec \theta d\theta$
Or we can also write it as
$\int {\ln \left( {\tan \theta + \sec \theta } \right).\sec \theta \tan \theta d\theta }$ ---------- (iii)
Let $\ln \left( {\tan \theta + \sec \theta } \right)$ be the first function and $\sec \theta \tan \theta$ be the second function. And we know that $\int {\sec \theta \tan \theta d\theta = \sec \theta + c}$ . On applying $uv$ formula of integration in equation (iii) we get,
$\ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta {\text{ }} - {\text{ }}\int {\left( {\dfrac{1}{{\tan \theta + \sec \theta }}.\left( {{{\sec }^2}\theta + \sec \theta \tan \theta } \right) \times \sec \theta } \right)} d\theta$
Take $\sec \theta$ common from $\left( {{{\sec }^2}\theta + \sec \theta \tan \theta } \right)$
$\Rightarrow \ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta {\text{ }} - {\text{ }}\int {\left( {\dfrac{{\sec \theta \left( {\sec \theta + \tan \theta } \right)}}{{\left( {\tan \theta + \sec \theta } \right)}}\sec \theta } \right)} d\theta$
and cancel out $\tan \theta + \sec \theta$ from numerator as well as from denominator
$\Rightarrow \ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta {\text{ }} - {\text{ }}\int {{{\sec }^2}\theta } d\theta$
$\because$ the value of $\int {{{\sec }^2}\theta } d\theta = \tan \theta + c$
$\Rightarrow \ln \left( {\tan \theta + \sec \theta } \right) \times \sec \theta {\text{ }} - \tan \theta + c$
$\because$ the value of $\tan \theta$ is $x$ and the value of $\sec \theta$ is $\sqrt {1 + {x^2}}$
$\Rightarrow \ln \left( {x + \sqrt {1 + {x^2}} } \right){\text{.}}\sqrt {1 + {x^2}} {\text{ }} - x + c$
We can also write it as
$\Rightarrow \sqrt {1 + {x^2}} .\ln \left( {x + \sqrt {1 + {x^2}} } \right){\text{ }} - x + c$

So, the correct answer is “Option 1”.

Note:
Remember that the value of ${\sec ^2}\theta$ is $1 + {x^2}$ by which the value of $\sec \theta$ is $\sqrt {1 + {x^2}}$ . Also the value of $\tan \theta$ is $x{\text{ }}$ . Also remember that ${\tan ^2}\theta + 1 = {\sec ^2}\theta$ and $\int {\sec \theta \tan \theta d\theta = \sec \theta + c}$ .If you are facing any problem regarding integration in question and linear equations in the given options then there are two ways to get the answer:
First, simply go with the theory and think a formula by using trigonometric functions like here we have seen that $1 + {x^2}$ is given which is related with $\tan x$ .So we have used $\tan x$ and $\sec x$ , $\sec x$ is a derivative of $\tan x$ . Therefore, by simply analyzing the questions pattern we can find the answer.
Second, And if you are good in derivative and poor in integration then differentiate each option and the type of equation you will get after differentiating is the expression given in the question then that option is the correct option.