{1}{{9+8 x-x²} d x=(x)+c then (x)=

Correct Answer is : sin⁡−1(x−45) ^{-1}≤ft(x-4/5)sin−1(5x−4)

Correct Answer is : sin⁡−1(x−45) ^{-1}≤ft(x-4/5)sin−1(5x−4)

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COMEDK 2025 Morning Solved Paper

Question : 137 of 180

Marks: +1, -0

\int \frac{1}{\sqrt{9+8 x-x^2}} d x=\varphi(x)+c

\varphi(x)=

Solution:

Concept:

The integral

\int \frac{dx}{\sqrt{a^2-(x-b)^2}}

equals

\sin^{-1}\left(\frac{x-b}{a}\right)+C

Explanation:

Rewrite the denominator by completing the square:

9+8x-x^2 = -(x^2-8x-9) = -[(x-4)^2-25] = 25-(x-4)^2

So the integral becomes

\int \frac{dx}{\sqrt{25-(x-4)^2}}

Let

u=x-4

and

a=5

. Then

\int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}\left(\frac{u}{a}\right)+C

Substitute back:

\varphi(x) = \sin^{-1}\left(\frac{x-4}{5}\right)

Answer:

\varphi(x) = \sin^{-1}\left(\frac{x-4}{5}\right)

(Option B).

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