Question

integrate x+1 /root of 8+x-x² with respect to x​

Answer 1

Step-by-step explanation:

We have,

$\int \frac{x + 1}{ \sqrt{8 + x - {x}^{2} } } dx$

$let \: \: 8 + x - {x}^{2} = {t}^{2} \\ \implies (- 2x + 1)dx = 2tdt$

$= - \frac{1}{2} \int \frac{ - 2x - 2}{ \sqrt{8 + x - {x}^{2} } } dx$

$= - \frac{1}{2} \int \frac{ (- 2x + 1) - 3}{ \sqrt{8 + x - {x}^{2} } } dx$

$= - \frac{1}{2} \int \frac{( - 2x + 1)dx}{ \sqrt{8 + x - {x}^{2} } } + \frac{3}{2} \int \frac{dx}{ \sqrt{8 + x - {x}^{2} } }$

$= - \frac{1}{2} \int \frac{2tdt}{t} + \frac{3}{2} \int \frac{dx}{ \sqrt{8 - ( {x}^{2} - x + \frac{1}{4} - \frac{1}{4} )} }$

$= - \int \: dt + \frac{3}{2} \int \frac{dx}{ \sqrt{8 + \frac{1}{4} - (x - \frac{1}{2} )^{2} } }$

$= - t + \frac{3}{2} \int\frac{dx}{ \sqrt{ \frac{33}{4} - (x - \frac{1}{2})^{2} } }$

$= - \sqrt{8 + x - {x}^{2} } + \frac{3}{2} \int \frac{dx}{ \sqrt{ {( \frac{ \sqrt{33} }{2}) }^{2} - (x - \frac{1}{2})^{2} } }$

$= - \sqrt{8 + x - {x}^{2} } + \frac{3}{2} . \frac{2}{ \sqrt{33} } \sin^{ - 1} ( \frac{x - \frac{1}{2} }{ \frac{ \sqrt{33} }{2} } )$

$= - \sqrt{8 + x - {x}^{2} } + \frac{3}{ \sqrt{33} } \sin^{ - 1} ( \frac{2x - 1}{ \sqrt{33} } ) + c$