Mathematics

Integrate $$\displaystyle \int {\sqrt {1 + {x^2}} } dx$$ is equal to:


SOLUTION
$$ = \mathop \smallint \nolimits^ {\sec ^2}\left( u \right)\sqrt {{{\tan }^2}\left( u \right) + 1} {\mkern 1mu} {\rm{d}}u$$
$$ = \mathop \smallint \nolimits^ {\sec ^3}\left( u \right){\mkern 1mu} {\rm{d}}u$$
 
 $$=\dfrac { { \sec  \left( u \right) \tan  \left( u \right)  } }{ 2 } +\dfrac { 1 }{ 2 } \int { sec\left( u \right)  } { { d } }u$$
 
$$\int {\sec \left( u \right)du} $$
$$ = \ln \left( {\tan \left( u \right) + \sec \left( u \right)} \right)$$
$$\dfrac{{\sec \left( u \right)\tan \left( u \right)}}{2} + \dfrac{1}{2}\int {\sec \left( u \right)du} $$
$$ = \dfrac{{\ln \left( {\tan \left( u \right) + \sec \left( u \right)} \right)}}{2} + \dfrac{{\sec \left( u \right)\tan \left( u \right)}}{2}$$
$$\mathop \smallint \nolimits^ \sqrt {{x^2} + 1} {\mkern 1mu} {\rm{d}}x$$

$$ = \dfrac{{\ln \left( {\left| {\sqrt {{x^2} + 1}  + x} \right|} \right)}}{2} + \dfrac{{x\sqrt {{x^2} + 1} }}{2} + C$$

$$\dfrac{{ar\sinh \left( x \right)}}{2} + \dfrac{{x\sqrt {{x^2} + 1} }}{2} + C$$

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Subjective Medium Published on 17th 09, 2020
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