Question

The integral $\displaystyle\int \frac{dx}{(1+ \sqrt{x}) \sqrt{x - x^2}}$ is equal to (where $C$ is a constant of integration)

• A. $-2 \sqrt{\frac{1+ \sqrt{x}}{1 - \sqrt{x}}} + C$
• B. $-2 \sqrt{\frac{1- \sqrt{x}}{1+ \sqrt{x}}} + C$
• C. $- \sqrt{\frac{1- \sqrt{x}}{1+\sqrt{x}}} + C$
• D. $2 \sqrt{\frac{1+ \sqrt{x}}{1 - \sqrt{x}}} + C$

Answer 1

The Correct Option is B

To solve the integral $\displaystyle\int \frac{dx}{(1+ \sqrt{x}) \sqrt{x - x^2}}$, we can use a substitution method. Let's start by analyzing the integrand and see which substitution would simplify it best. Notice that the expression under the square root, $x - x^2$, resembles the standard form of a perfect square trinomial when manipulated: $x - x^2 = x(1-x)$.

To tackle the integral, we make the substitution $x = \sin^2 \theta$, which implies $dx = 2 \sin \theta \cos \theta \, d\theta = \sin(2\theta) \, d\theta$. This choice turns $x - x^2 = \sin^2 \theta - \sin^4 \theta = \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta$. The square root of this expression leads to a simplification in terms of trigonometric identities.

Substituting these into the integral gives:

$$ \int \frac{\sin(2\theta) \, d\theta}{(1 + \sin \theta) \frac{1}{2} \sin 2\theta} = \int \frac{2 \, d\theta}{1 + \sin \theta}. $$

We simplify further by using the identity $1 + \sin \theta = 2 \sin^2\left(\frac{\theta}{2}\right)$. This implies:

$$ \int \frac{2 \, d\theta}{2 \sin^2\left(\frac{\theta}{2}\right)} = \int \csc^2 \left(\frac{\theta}{2}\right) d\theta. $$

Recognizing this as a standard trigonometric integral, we know the antiderivative of $\csc^2 x$ is $-\cot x$. Therefore:

$$ -2 \cot\left(\frac{\theta}{2}\right) + C. $$

Reversing our substitution requires us to express $\theta$ in terms of $x$. Since $x = \sin^2 \theta$, we have $ \sin \theta = \sqrt{x}$, and thus $ \theta = \sin^{-1}(\sqrt{x})$. Then:

$$ \cot\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} = \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}. $$

Thus, the integral evaluates to:

$$ -2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} + C. $$

Hence, the correct answer is $-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} + C$.