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

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Question

(A) Let $I = \int \frac{6x+5}{\sqrt{6+x-2x^2}} dx$. We express the numerator as $6x+5 = A \frac{d}{dx}(6+x-2x^2) + B$. $6x+5 = A(1-4x) + B = -4Ax + (A

Vedclass · Accepted Answer

$-3 \sqrt{6+x-2x^2} + \frac{13}{2\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right) + c$

Vedclass · Answer

(A) Let $I = \int \frac{6x+5}{\sqrt{6+x-2x^2}} dx$. We express the numerator as $6x+5 = A \frac{d}{dx}(6+x-2x^2) + B$. $6x+5 = A(1-4x) + B = -4Ax + (A+B)$. Comparing coefficients,$-4A = 6 \implies A = -\frac{3}{2}$ and $A+B = 5 \implies B = 5 + \frac{3}{2} = \frac{13}{2}$. So,$I = \int \frac{-\frac{3}{2}(1-4x) + \frac{13}{2}}{\sqrt{6+x-2x^2}} dx = -\frac{3}{2} \int \frac{1-4x}{\sqrt{6+x-2x^2}} dx + \frac{13}{2} \int \frac{dx}{\sqrt{6+x-2x^2}}$. For the first part,let $u = 6+x-2x^2$,then $du = (1-4x)dx$. Integral becomes $-\frac{3}{2} \int u^{-1/2} du = -\frac{3}{2} (2\sqrt{u}) = -3\sqrt{6+x-2x^2}$. For the second part,$\int \frac{dx}{\sqrt{6+x-2x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{3 + \frac{x}{2} - x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{49}{16} - (x^2 - \frac{x}{2} + \frac{1}{16})}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{(\frac{7}{4})^2 - (x-\frac{1}{4})^2}}$. This is $\frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{x-1/4}{7/4}\right) = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right)$. Combining both,$I = -3\sqrt{6+x-2x^2} + \frac{13}{2\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right) + c$.

$\int \frac{6x+5}{\sqrt{6+x-2x^2}} dx =$

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