{dx}{{9-8x-4x²} = . . . . . . + C

Correct Answer is : 12sin⁡−1(2x+213)1/2 ^{-1} ({2x+2}{{13}})21sin−1(132x+2)

Correct Answer is : 12sin⁡−1(2x+213)1/2 ^{-1} ({2x+2}{{13}})21sin−1(132x+2)

Test Index

GUJCET 2026 Math Paper

Question : 5 of 40

Marks: +1, -0

\int \frac{dx}{\sqrt{9-8x-4x^2}} =

C

Solution:

Concept:

Integrals of the form

\int \frac{dx}{\sqrt{a^2-u^2}}

give an inverse sine function.

Explanation:

Let

I=\int \frac{dx}{\sqrt{9-8x-4x^2}}

Complete the square in the denominator:

9-8x-4x^2=9-4(x^2+2x)

=9-4[(x+1)^2-1]

=13-4(x+1)^2

=13-(2x+2)^2

So,

I=\int \frac{dx}{\sqrt{13-(2x+2)^2}}

Put

u=2x+2

. Then

du=2\,dx

, so

dx=\frac{du}{2}

Thus,

I=\frac{1}{2}\int \frac{du}{\sqrt{(\sqrt{13})^2-u^2}}

Using

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

I=\frac{1}{2}\sin^{-1}\left(\frac{u}{\sqrt{13}}\right)+C

Substitute

u=2x+2

I=\frac{1}{2}\sin^{-1}\left(\frac{2x+2}{\sqrt{13}}\right)+C

Answer:

\frac{1}{2}\sin^{-1}\left(\frac{2x+2}{\sqrt{13}}\right)+C

Hence, option D is correct.

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