int frac{2x+5}{sqrt{7-6x-x^2}} , dx = A sqrt{ | vedclass

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(A) Let $I = \int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx$.We rewrite the numerator as $2x+5 = -( -2x - 6 ) - 1$.So,$I = \int \frac{-( -2x - 6 ) - 1}{\sqrt

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-$3$

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(A) Let $I = \int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx$.We rewrite the numerator as $2x+5 = -( -2x - 6 ) - 1$.So,$I = \int \frac{-( -2x - 6 ) - 1}{\sqrt{7-6x-x^2}} \, dx = -\int \frac{-2x-6}{\sqrt{7-6x-x^2}} \, dx - \int \frac{1}{\sqrt{7 - (x^2 + 6x)}} \, dx$.Completing the square in the denominator: $7 - (x^2 + 6x + 9 - 9) = 7 + 9 - (x+3)^2 = 16 - (x+3)^2 = 4^2 - (x+3)^2$.Thus,$I = -\int (7-6x-x^2)^{-1/2} (-2x-6) \, dx - \int \frac{1}{\sqrt{4^2 - (x+3)^2}} \, dx$.Using the substitution $u = 7-6x-x^2$,$du = (-6-2x) \, dx$,we get:$I = -2(7-6x-x^2)^{1/2} - \sin^{-1}\left(\frac{x+3}{4}\right) + c$.Comparing this with the given form $A \sqrt{7-6x-x^2} + B \sin^{-1}\left(\frac{x+3}{4}\right) + c$,we find $A = -2$ and $B = -1$.Therefore,$A+B = -2 + (-1) = -3$.

$\int \frac{2x+5}{\sqrt{7-6x-x^2}} \, dx = A \sqrt{7-6x-x^2} + B \sin^{-1}\left(\frac{x+3}{4}\right) + c$ (where $c$ is a constant of integration),then the value of $A+B$ is

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