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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