int frac{d x}{sqrt{5+4 x-x^{2}}}= | vedclass

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Question

(A) To evaluate the integral $I = \int \frac{dx}{\sqrt{5+4x-x^2}}$,we first complete the square for the quadratic expression inside the square root.$5

Vedclass · Accepted Answer

$\sin ^{-1}\left(\frac{x-2}{3}\right)+c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{dx}{\sqrt{5+4x-x^2}}$,we first complete the square for the quadratic expression inside the square root.$5+4x-x^2 = -(x^2-4x-5) = -((x-2)^2 - 4 - 5) = -( (x-2)^2 - 9 ) = 9 - (x-2)^2$.Substituting this back into the integral,we get:$I = \int \frac{dx}{\sqrt{3^2 - (x-2)^2}}$.Using the standard integration formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we obtain:$I = \sin^{-1}\left(\frac{x-2}{3}\right) + c$.

$\int \frac{d x}{\sqrt{5+4 x-x^{2}}}=$

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