Question

$\int\limits^4_0 dx/√12+4x-x² ,Evaluate the given integral expression:$

Answer 1

HELLO DEAR,



GIVEN:-
$\sf{\int\limits^4_0 dx/(\sqrt{12} + 4x - x^2)}$

= $\sf{\int\limits^4_0 dx/[-(x^2 - 4x + 4 - 4 - \sqrt{12})]}$

= $\sf{\int\limits^4_0 dx/[(4 + \sqrt{12}) - (x - 2)^2]}$

= $\sf{\int\limits^4_0 dx/[(1 + 3 + 2\sqrt{3}) - (x - 2)^2]}$

= $\sf{\int\limits^4_0 dx/[(1 + \sqrt{3})^2 - (x - 2)^2]}$

= $\sf{[1/2(1 + \sqrt{3}) log|{{(1 + \sqrt{3} + (x - 2)}\over {(1 + \sqrt{3} - (x - 2)}}|]^4_0}$

= $\sf{[1/2(1 + \sqrt{3})[log|{(\sqrt{3} - 1 + 4)\over(3 + \sqrt{3} - 4)}| - log|{(\sqrt{3} - 1 + 0)\over(3 + \sqrt{3} - 0)}|]}$

= 1/2(1 - √3) $\sf[{log|{(\sqrt{3} + 3)\over(\sqrt{3} - 1)}| - log|{(\sqrt{3} - 1)\over(\sqrt{3} + 4)}|]}$


I HOPE ITS HELP YOU,
THANKS