x . , , , ( x , , + , , 1 , + , x^2 ) 1 , + , x^2 , dx equals : - Vidyadip

MCQ

$\int x .\frac{{\ln \,\,\,\left( {x\,\, + \,\,\sqrt {1\, + \,{x^2}} } \right)}}{{\sqrt {1\, + \,{x^2}} }} \, dx$ equals :

✓
$\sqrt {1 + {x^2}} $ $ln$ $\left( {x\,\, + \,\,\sqrt {1\, + \,{x^2}} } \right)$ $- x + c$
B
$\frac{x}{2}$. $ln^2$ $\left( {x\,\, + \,\,\sqrt {1\, + \,{x^2}} } \right)$ $- \frac{x}{{\sqrt {1\,\, + \,\,{x^2}} }}$ $+ c$
C
$\frac{x}{2}$. $ln^2$ $\left( {x\,\, + \,\,\sqrt {1\, + \,{x^2}} } \right)$ $+ \frac{x}{{\sqrt {1\,\, + \,\,{x^2}} }}$ $+ c$
D
$\sqrt {1 + {x^2}} $ $ln$ $\left( {x\,\, + \,\,\sqrt {1\, + \,{x^2}} } \right)$ $ +x + c$

✓

Answer

Correct option: A.

$\sqrt {1 + {x^2}} $ $ln$ $\left( {x\,\, + \,\,\sqrt {1\, + \,{x^2}} } \right)$ $- x + c$

a
use $I.B.P$. taking $ln$ $\left( {x + \sqrt {1 + {x^2}} } \right)$ as the first function and $\frac{x}{{\sqrt {1\,\, + \,\,{x^2}} }}$ as the second function

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