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

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Question

(A) ધારો કે $I = \int \frac{6x+5}{\sqrt{6+x-2x^2}} dx$. અંશને $6x+5 = A \frac{d}{dx}(6+x-2x^2) + B$ તરીકે દર્શાવીએ. $6x+5 = A(1-4x) + B = -4Ax + (A+B)

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) ધારો કે $I = \int \frac{6x+5}{\sqrt{6+x-2x^2}} dx$. અંશને $6x+5 = A \frac{d}{dx}(6+x-2x^2) + B$ તરીકે દર્શાવીએ. $6x+5 = A(1-4x) + B = -4Ax + (A+B)$. સહગુણકોની સરખામણી કરતા,$-4A = 6 \implies A = -\frac{3}{2}$ અને $A+B = 5 \implies B = 5 + \frac{3}{2} = \frac{13}{2}$. તેથી,$I = \int \frac{-\frac{3}{2}(1-4x) + \frac{13}{2}}{\sqrt{6+x-2x^2}} dx = -\frac{3}{2} \int \frac{1-4x}{\sqrt{6+x-2x^2}} dx + \frac{13}{2} \int \frac{dx}{\sqrt{6+x-2x^2}}$. પ્રથમ ભાગ માટે,$u = 6+x-2x^2$ લેતા,$du = (1-4x)dx$ મળે. સંકલન $-\frac{3}{2} \int u^{-1/2} du = -\frac{3}{2} (2\sqrt{u}) = -3\sqrt{6+x-2x^2}$ થાય. બીજા ભાગ માટે,$\int \frac{dx}{\sqrt{6+x-2x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{3 + \frac{x}{2} - x^2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{\frac{49}{16} - (x-\frac{1}{4})^2}} = \frac{1}{\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right)$. બંનેને જોડતા,$I = -3\sqrt{6+x-2x^2} + \frac{13}{2\sqrt{2}} \sin^{-1}\left(\frac{4x-1}{7}\right) + c$ મળે.

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

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