Question

Find the following integrals :-

$(1). \: \int \frac{dx}{\ \textless \ br /\ \textgreater \ {x}^{2} - 16} \\ \\ (2). \: \int \frac{dx}{ \sqrt{2x - {x}^{2} } }$
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Answer 1

$\begin{gathered}\Large{\bold{\green{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \rm :\implies\: \boxed{ \pink{ \bf \: \int\dfrac{dx}{ {x}^{2} - {a}^{2} } \: = \tt \: \dfrac{1}{2a} log(\dfrac{x - a}{x + a} ) \: + \: c}}$

$(2). \: \rm :\implies\: \boxed{ \pink{ \bf \: \int \: \dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } \: = \tt \: {sin}^{ - 1} \dfrac{x}{a} \: + \: c }}$

$\large\underline\purple{\bold{Solution - }}$

$\bf \: (1). \: \int \: \dfrac{dx}{ {x}^{2} - 16}$

$\rm :\implies\:\int \: \dfrac{dx}{ {x}^{2} - {4}^{2} }$

$\rm :\implies\:\dfrac{1}{2 \times 4} \: log(\dfrac{x \: - \: 4}{x \: + \: 4} ) \: + \: c$

$\rm :\implies\:\dfrac{1}{8} \: log(\dfrac{x \: - \: 4}{x \: + \: 4} ) \: + \: c$

$\rm :\implies\: \boxed{ \green{ \bf \: \int \: \dfrac{dx}{ {x}^{2} - 16} \: = \tt \:\dfrac{1}{8} log(\dfrac{x - 4}{x + 4} ) \: + \: c }}$

$\bf \: (2). \: \int \: \dfrac{dx}{ \sqrt{2x - {x}^{2} } }$

$\rm :\implies\:\int \dfrac{dx}{ \sqrt{-({x}^{2} - 2x)} }$

$\rm :\implies\:\int \dfrac{dx}{ \sqrt{ - ( {x}^{2} - 2x + 1 - 1) } }$

$\rm :\implies\: \int \: \dfrac{dx}{ \sqrt{ - \bigg( {(x - 1)}^{2} - 1 \bigg) } }$

$\rm :\implies\: \int \: \dfrac{dx}{ \sqrt{1 - {(x - 1)}^{2} } }$

$\rm :\implies\: {sin}^{ - 1} \dfrac{x - 1}{1} \: + \: c$

$\rm :\implies\: {sin}^{ - 1} (x - 1) + c$

$\rm :\implies\: \boxed{ \pink{ \bf \: \int \dfrac{dx}{ \sqrt{2x - {x}^{2} } } \: = \tt \: {sin}^{ - 1} (x - 1) \: + \: c}}$