Solveeit Logo
Login

Question

Question: \(\int_{}^{}{\frac{dx}{\sqrt{2 - 3x - x^{2}}} = fog(x) + C}\), then...

∫dx2−3x−x2=fog(x)+C\int_{}^{}{\frac{dx}{\sqrt{2 - 3x - x^{2}}} = fog(x) + C}, then

A

f(x)=sin⁡−1x,g(x)=(2x−3)17f(x) = \sin^{- 1}x,g(x) = \frac{(2x - 3)}{\sqrt{17}}

B

f(x)=tan⁡−1x,g(x)=(2x+3)17f(x) = \tan^{- 1}x,g(x) = \frac{(2x + 3)}{\sqrt{17}}

C

f(x)=sin⁡−1x,g(x)=(2x+3)17f(x) = \sin^{- 1}x,g(x) = \frac{(2x + 3)}{\sqrt{17}}

D

None of these

Answer

f(x)=sin⁡−1x,g(x)=(2x+3)17f(x) = \sin^{- 1}x,g(x) = \frac{(2x + 3)}{\sqrt{17}}

Explanation

Solution

We have 2−3x−x2=−(x2+3x−2)2 - 3x - x^{2} = - \left( x^{2} + 3x - 2 \right)

=−[(x+32)2−174]- \left[ \left( x + \frac { 3 } { 2 } \right) ^ { 2 } - \frac { 17 } { 4 } \right]∴ Therefore, g(x)=(2x+3)17g(x) = \frac{(2x + 3)}{\sqrt{17}}and f(x)=sin⁡−1xf(x) = \sin^{- 1}x.