Question

$\bigstar\sf{Integrate}$

${\displaystyle{\int × \sqrt{\dfrac{a^2-x^2}{a^2+x^2}}}}$​

Answer 1

Answer:

here is answer dear friend

Answer 2

SOLUTION:-

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We have

$\displaystyle \bf \int \dfrac{ \sqrt{x^{2} + a^{2} } }{x}$

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$\implies\displaystyle \sf \int \dfrac{ \sqrt{x^{2} + a^{2} } \sqrt{x^{2} + a ^{2} } }{x \sqrt{x {}^{2} + a ^{2} } } </p><p>$

$\implies \displaystyle \sf \int \dfrac{x^{2} + {a}^{2} }{x \sqrt{ {x}^{2}+{a}^{2}} }$

⠀⠀

$\implies \displaystyle \sf \int \dfrac{x^{2} dx}{x \sqrt{ {x}^{2} + {a}^{2} } } + \int\dfrac{a^{2} dx}{x \sqrt{ {x}^{2} + {a}^{2} }}$

$\implies \displaystyle \sf \dfrac{1}{2} \int(2x )( {x}^{2} + {a}^{2} )^{ - \frac{1 }{2} } dx + {a}^{2} \int \frac{dx}{x \sqrt{ {x}^{2} + {a}^{2} } }$

$\implies \displaystyle \sf \frac{1}{2} \bigg \{ \sf \dfrac{( {x}^{2} + {a}^{2} ) ^{ \frac{1}{2} } }{ \dfrac{1}{2} } \bigg \} + a^{2} I_{1} + c....(1)$

Where

$\displaystyle \sf I_{1} = \int \dfrac{dx}{x \sqrt{ {x}^{2} + {a}^{2} } }$

✶Put x=1/t ⇛dx=-1 /t² dt

$\implies \displaystyle \sf \int \frac{ - \dfrac{1}{{t}^{2}}dt }{ \dfrac{1}{t} \sqrt{ \dfrac{1 }{ {t}^{2}} + {a}^{2} } }$

$\implies \displaystyle \sf \int \dfrac{ - dt}{ \sqrt{1 + {a}^{2} {t}^{2} } }$

$\implies \displaystyle \sf \frac{1}{a} \int \dfrac{dt}{ \sqrt{ {t}^{2} + \dfrac{1}{ {a}^{2} } } }</p><p>$

$\implies \displaystyle \sf - \frac{1}{a} log \bigg(t + \sqrt{ {t}^{2} + \dfrac{1}{ {a}^{2} } } \bigg)$

$\implies \displaystyle \sf - \frac{1}{a} log \bigg( \frac{1}{x} + \sqrt{ \frac{1}{ {x}^{2} } + \dfrac{1}{ {a}^{2} } } \bigg)</p><p>$

$\implies \displaystyle \sf - \frac{1}{a} log \bigg( \frac{a + \sqrt{ {x}^{2} + {a}^{2} } }{ax} \bigg)....</p><p>$

⠀⠀⠀⠀

From (1) and (2) we find that given integral

$\implies \bf\sqrt{ {x}^{2} + {a}^{2} } - a \: log \bigg( \dfrac{a + \sqrt{ {x}^{2} + {a}^{2} } }{xa} \bigg)$