Mathematics

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

##### SOLUTION
$\int {x.\,\frac{{l\,n\left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {1 + {x^2}} }}}$

$\int {l\,n\,\left( {x + \sqrt {1 + {x^2}} } \right)} \,.\,\frac{1}{{\sqrt {1 + {x^2}} }}$

$u = l\,n\,\left( {x + \sqrt {1 + {x^2}} } \right)$

$dv\, = \frac{1}{{x + \sqrt {1 + {x^2}} }}\left( {1 + \frac{1}{{2\sqrt {1 + {x^2}} }} \times 2x} \right)$

$dv = \frac{1}{{x + \sqrt {1 + {x^2}} }}\left( {1 + \frac{{\sqrt {1 + {x^2}} + x}}{{\sqrt {1 + {x^2}} }}} \right)$

$dv = \,\frac{1}{{\sqrt {1 + {x^2}} }}dx$

$\Rightarrow \int {udv = \frac{{{u^2}}}{2}}$

$\Rightarrow \frac{{l\,n\,\left( {x + \sqrt {1 + {x^2}} } \right)}}{2} + c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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