Question

Intrigation (1-x) dx by (1+x) root over x+x²+x³

Answer 1

Step-by-step explanation:

We have,

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

$let \: \: x = t^{2} \\ \implies \: dx = 2tdt$

$= \int \frac{(1 - {t}^{2}).2tdt }{(1 + {t}^{2} ) \sqrt{ {t}^{2} + {t}^{4} + {t}^{6} } }$

$= 2 \int \frac{(1 - {t}^{2})tdt }{(1 + {t}^{2}).t \sqrt{1 + {t}^{2} + {t}^{4} } }$

$= 2 \int \frac{(1 - {t}^{2})dt }{t( \frac{1}{t} + t)t \sqrt{ 1 + \frac{1}{ {t}^{2} } + {t}^{2} } }$

$= - 2 \int \frac{(1 - \frac{1}{ {t}^{2} } )dt }{( \frac{1}{t} + t)\sqrt{ 1 + \frac{1}{ {t}^{2} } + {t}^{2} } }$

$= - 2 \int \frac{(1 - \frac{1}{ {t}^{2} } )dt }{( \frac{1}{t} + t)\sqrt{ (t + \frac{1}{t} )^{2} - 1 } }$

$let \: \: t + \frac{1}{t} = u \\ \implies(1 - \frac{1}{ {t}^{2} } )dt = du$

$= - 2 \int \frac{du}{ u \sqrt{ {u}^{2} - 1 } }$

$= - 2 \sec^{ - 1} (u) + c$

$= - 2\sec^{ - 1} (t + \frac{1}{t} ) + c$

$= - 2\sec^{ - 1} ( \sqrt{x} + \frac{1}{ \sqrt{x} } ) + c$