Question

$\\ \odot \: \: \: \: \: \: \: \large \bf \underline{ \underline{Solve \: \: it} \: } _{ \bigstar \star}$
$\dashrightarrow \: \: \sf \int \frac{(x - 1)}{(x + 1) \sqrt{ {x}^{3} + {x}^{2} + x} } dx$
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Answer 1

SOLUTION :

$\: \: \: \: \: \: \int \frac{(x - 1)}{(x + 1) \sqrt{ {x}^{3} + {x}^{2} + x } } dx$

$\longrightarrow \int \frac{({ {x}^{2} - 1)}^{} }{ {(x + 1)}^{2} \sqrt{ {x}^{3} + {x}^{2} + x} } dx$

$\longrightarrow \int \frac{(1 - \frac{1}{ {x}^{2} }) }{ \frac{ {x}^{2} + 2x + 1}{x} \sqrt{x + 1 + \frac{1}{x} } } dx$

$\longrightarrow \int\frac{(1 - \frac{1}{ {x}^{2} }) }{ { \bigg({x}^{} + 2 + \frac{1}{x} \bigg)}{} \sqrt{x + 1 + \frac{1}{x} } } dx$

$Let, \\ x + 2 + 1/x = z \\ \\ \longrightarrow \bigg(1 + \frac{1}{ {x}^{2} } \bigg) dx = 2z \: dz$

Now,

$\longrightarrow \int\frac{(1 - \frac{1}{ {x}^{2} }) }{ { \bigg({x}^{} + 2 + \frac{1}{x} \bigg)}{} \sqrt{x + 1 + \frac{1}{x} } } dx = \int \frac{2z }{( {z}^{2} + 1)z} dz\\ \\ \\ \longrightarrow \: 2 \int \frac{dz}{ {z}^{2} + 1 } \\ \\ \\ \longrightarrow \: 2 { \tan}^{ - 1} (z) + c \\ \\ \\ \longrightarrow \bf \: 2 { \tan}^{ - 1} \bigg(x + 2 + \frac{1}{x} \bigg) + c$