Question

31 question DX upon x-1 under root x squarec+ 4​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{dx}{(x - 1) \sqrt{ {x}^{2} + 4 } }$

$let \: \: x = 2 \tan( \alpha ) \\ \implies \: dx = 2 \sec^{2} ( \alpha ) d \alpha$

$= \int \frac{2 \sec^{2} ( \alpha )d \alpha }{(2 \tan( \alpha ) - 1).2 \sec( \alpha ) }$

$= \int \frac{ \sec( \alpha )d \alpha }{2 \tan( \alpha ) - 1 }$

$= \int \frac{d \alpha }{2 \sin( \alpha ) - \cos( \alpha ) }$

$= \int \frac{d \alpha }{ \frac{4 \tan( \frac{ \alpha }{2} ) }{1 + \tan^{2} ( \frac{ \alpha }{2} ) } - \frac{1 - \tan^{2} ( \frac{ \alpha }{2} ) }{1 + \tan^{2} ( \frac{ \alpha }{2} ) } }$

$= \int \frac{ \sec^{2} ( \frac{ \alpha }{2} ) d \alpha }{ \tan^{2} ( \frac{ \alpha }{2} ) + 4 \tan( \frac{ \alpha }{2} ) - 1 }$

$let \: \: \tan( \frac{ \alpha }{2} ) = t \\ \implies \sec^{2} ( \frac{ \alpha }{2} ) d \alpha = 2dt$

$= \int \frac{2dt}{ {t}^{2} + 4t - 1}$

$= 2 \int \frac{dt}{ {t}^{2} + 4t + 4 - 5 }$

$= 2 \int \frac{dt}{(t + 2)^{2} - ( \sqrt{5})^{2} }$

$= 2( \frac{1}{2 \sqrt{5} } ln |\frac{t + 2 - \sqrt{5} }{t + 2 + \sqrt{5} }| ) + c$

$= \frac{1}{ \sqrt{5} } ln | \frac{ \tan( \frac{ \alpha }{2} ) + 2 - \sqrt{5} }{ \tan( \frac{ \alpha }{2} ) + 2 + \sqrt{5} } | + c$

$= \frac{1}{ \sqrt{5} } ln | \frac{ \tan( \frac{ 1 }{2} \tan^{ - 1} ( \frac{x}{2} ) ) + 2 - \sqrt{5} }{ \tan( \frac{ 1 }{2} \tan^{ - 1} ( \frac{x}{2} ) ) + 2 + \sqrt{5} } | + c$