Question

integral √(1+x^2) dx​

Answer 1

Step-by-step explanation:

$i = \int \sqrt{1 + {x}^{2} } dx$

$i = \int \sqrt{1 + {x}^{2} } .1dx$

Now, by integration by parts we have,

$i= \sqrt{1 + {x}^{2} } \int1dx - \int( \frac{d}{dx} ( \sqrt{1 + {x}^{2} } ) \int1dx)dx$

$i = x \sqrt{1 + {x}^{2} } - \int \frac{2x}{2 \sqrt{1 + {x}^{2} } }.xdx$

$i = x \sqrt{1 + {x}^{2} } - \int \frac{1 + {x}^{2} - 1}{ \sqrt{1 + {x}^{2} } } dx$

$i = x \sqrt{1 + {x}^{2} } - \int \sqrt{1 + {x}^{2} } dx + \int \frac{1}{ \sqrt{1 + {x}^{2} } } dx$

$i = x \sqrt{1 + {x}^{2} } - i + \tan^{ - 1} (x)$

$\implies2i = x \sqrt{1 + {x}^{2} } + \tan^{ - 1} (x)$

$\implies \: i = \frac{x}{2} \sqrt{1 + {x}^{2} } + \frac{1}{2} \tan^{ - 1} (x)$

So,

the required integral is

$\frac{x}{2} \sqrt{1 + {x}^{2} } + \frac{1}{2} \tan^{ - 1} (x)$