Q106 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
$\int \dfrac{2 x}{\sqrt{1-x^2-x^4}} d x=\sin ^{-1}[f(x)]+c$ then $f(x)=$
- A$\dfrac{2 x^2}{\sqrt{3}}$
- B$\dfrac{2 x^2-1}{\sqrt{3}}$
- C$\dfrac{2 x^2}{\sqrt{5}}$
- Correct answerD$\dfrac{2 x^2+1}{\sqrt{5}}$
Solution
Let $I = \int \dfrac{2x}{\sqrt{1 - x^2 - x^4}} dx$.
Substitute $t = x^2$, so $dt = 2x dx$.
The integral becomes $I = \int \dfrac{dt}{\sqrt{1 - t - t^2}}$.
Complete the square in the denominator: $1 - t - t^2 = 1 - (t^2 + t) = 1 - (t^2 + t + \dfrac{1}{4} - \dfrac{1}{4}) = 1 + \dfrac{1}{4} - (t + \dfrac{1}{2})^2 = \dfrac{5}{4} - (t + \dfrac{1}{2})^2$.
Thus, $I = \int \dfrac{dt}{\sqrt{(\dfrac{\sqrt{5}}{2})^2 - (t + \dfrac{1}{2})^2}}$.
Using the standard integral $\int \dfrac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\dfrac{u}{a}) + C$, we get $I = \sin^{-1}(\dfrac{t + 1/2}{\sqrt{5}/2}) + C$.
Simplifying the argument: $\dfrac{t + 1/2}{\sqrt{5}/2} = \dfrac{2t + 1}{\sqrt{5}}$.
Substituting $t = x^2$ back, $I = \sin^{-1}(\dfrac{2x^2 + 1}{\sqrt{5}}) + C$.
Comparing with $\sin^{-1}[f(x)] + C$, we find $f(x) = \dfrac{2x^2 + 1}{\sqrt{5}}$.
Answer: $\dfrac{2 x^2+1}{\sqrt{5}}$
Substitute $t = x^2$, so $dt = 2x dx$.
The integral becomes $I = \int \dfrac{dt}{\sqrt{1 - t - t^2}}$.
Complete the square in the denominator: $1 - t - t^2 = 1 - (t^2 + t) = 1 - (t^2 + t + \dfrac{1}{4} - \dfrac{1}{4}) = 1 + \dfrac{1}{4} - (t + \dfrac{1}{2})^2 = \dfrac{5}{4} - (t + \dfrac{1}{2})^2$.
Thus, $I = \int \dfrac{dt}{\sqrt{(\dfrac{\sqrt{5}}{2})^2 - (t + \dfrac{1}{2})^2}}$.
Using the standard integral $\int \dfrac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\dfrac{u}{a}) + C$, we get $I = \sin^{-1}(\dfrac{t + 1/2}{\sqrt{5}/2}) + C$.
Simplifying the argument: $\dfrac{t + 1/2}{\sqrt{5}/2} = \dfrac{2t + 1}{\sqrt{5}}$.
Substituting $t = x^2$ back, $I = \sin^{-1}(\dfrac{2x^2 + 1}{\sqrt{5}}) + C$.
Comparing with $\sin^{-1}[f(x)] + C$, we find $f(x) = \dfrac{2x^2 + 1}{\sqrt{5}}$.
Answer: $\dfrac{2 x^2+1}{\sqrt{5}}$
Hence, the correct answer is option: 'D'.