int {frac{{2x + 5}}{{sqrt {7 - 6x - {x^2}} }} | vedclass

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Question

(A) To solve the integral $I = \int \frac{2x + 5}{\sqrt{7 - 6x - x^2}} dx$,we express the numerator in terms of the derivative of the quadratic expres

Vedclass · Accepted Answer

$(-2, -1)$

Vedclass · Answer

(A) To solve the integral $I = \int \frac{2x + 5}{\sqrt{7 - 6x - x^2}} dx$,we express the numerator in terms of the derivative of the quadratic expression inside the square root.Let $f(x) = 7 - 6x - x^2$. Then $f'(x) = -6 - 2x$.We write $2x + 5 = -( -2x - 6 ) - 1$.Thus,$I = \int \frac{-( -2x - 6 ) - 1}{\sqrt{7 - 6x - x^2}} dx = -\int \frac{-2x - 6}{\sqrt{7 - 6x - x^2}} dx - \int \frac{1}{\sqrt{7 - 6x - x^2}} dx$.For the first integral,let $u = 7 - 6x - x^2$,so $du = (-6 - 2x) dx$. Then $\int \frac{du}{\sqrt{u}} = 2\sqrt{u} = 2\sqrt{7 - 6x - x^2}$.So,$-\int \frac{-2x - 6}{\sqrt{7 - 6x - x^2}} dx = -2\sqrt{7 - 6x - x^2}$.For the second integral,complete the square: $7 - 6x - x^2 = 16 - (x^2 + 6x + 9) = 4^2 - (x + 3)^2$.So,$\int \frac{1}{\sqrt{4^2 - (x + 3)^2}} dx = \sin^{-1}\left(\frac{x + 3}{4}\right)$.Combining these,$I = -2\sqrt{7 - 6x - x^2} - \sin^{-1}\left(\frac{x + 3}{4}\right) + C$.Comparing with $A\sqrt{7 - 6x - x^2} + B\sin^{-1}\left(\frac{x + 3}{4}\right) + C$,we get $A = -2$ and $B = -1$.

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

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