Let α 0. If ∫₀ᵅ x / (√(x+α) - √x) dx = (16+20√2)/15, then α is equal to : (1) 2, (2) 4, (3) √2, (4

Q: Why do we rationalize the denominator? Can't we just integrate it as it is?

Great question! Integrating $\frac{x}{\sqrt{x+\alpha} - \sqrt{x}}$ directly is extremely difficult because the variable $x$ is trapped inside the radical difference in the denominator. Rationalization is a powerful algebraic tool that 'lifts' the variables out of the denominator, turning a complex radical expression into a simple constant (in this case, $\alpha$), which makes the subsequent integration straightforward.

Q: When I substitute $t = x+\alpha$, do I really need to change the limits of integration?

Yes, absolutely! Forgetting to change the limits is the most common source of errors in definite integrals. When you change the variable from $x$ to $t$, the boundaries of the integration must also change to reflect the new variable. For $t = x+\alpha$, when $x=0$, $t=\alpha$, and when $x=\alpha$, $t=2\alpha$. Always update your limits immediately after choosing your substitution!

Q: Is there a shortcut for $\int x\sqrt{x+\alpha} dx$?

While you could use integration by parts, substitution is the most robust and elegant method here. By setting $t = x+\alpha$, you transform the integrand into $(t-\alpha)\sqrt{t}$, which is just a simple polynomial in $t$. This avoids the complexity of parts and reduces the chance of algebraic errors.

Question

Let $\alpha > 0$. If $\int_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha} - \sqrt{x}} dx = \frac{16+20\sqrt{2}}{15}$, then $\alpha$ is equal to :

Options:

(A) $2$

(B) $4$

(C) $\sqrt{2}$

(D) $2\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of JEE mastery! Today, we are standing before a beautiful, albeit intimidating, definite integral.

We are given the integral:

$$I = \int_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha} - \sqrt{x}} dx$$

We are told that this integral equals $\frac{16+20\sqrt{2}}{15}$. Our mission is to find the value of $\alpha$.

### The Art of Rationalization

Whenever you see a denominator involving the difference of two square roots, like $\sqrt{x+\alpha} - \sqrt{x}$, your mathematical intuition should immediately scream **"Rationalize!"** It is a classic trap to try and integrate this directly.

Instead, we multiply the numerator and the denominator by the conjugate, $(\sqrt{x+\alpha} + \sqrt{x})$.

In the denominator, we apply the identity $(a-b)(a+b) = a^2 - b^2$. The denominator becomes:

$$(\sqrt{x+\alpha})^2 - (\sqrt{x})^2 = (x+\alpha) - x = \alpha$$

Just like that, the radical disappears from the denominator, leaving us with a constant $\alpha$. Our integrand now simplifies to:

$$\frac{x(\sqrt{x+\alpha} + \sqrt{x})}{\alpha}$$

### Divide and Conquer

With $\alpha$ as a constant, we can pull $\frac{1}{\alpha}$ outside the integral. We then distribute the $x$ in the numerator to get two separate terms: $x\sqrt{x+\alpha}$ and $x^{3/2}$.

We can now split our integral into two bite-sized pieces:

$$I = \frac{1}{\alpha} \left[ \int_{0}^{\alpha} x\sqrt{x+\alpha} dx + \int_{0}^{\alpha} x^{3/2} dx \right]$$

### The Substitution

The second integral, $\int_{0}^{\alpha} x^{3/2} dx$, is a straightforward application of the power rule. It evaluates to $\frac{2}{5}\alpha^{5/2}$.

For the first part, $\int_{0}^{\alpha} x\sqrt{x+\alpha} dx$, we use the substitution $t = x+\alpha$. Then $dt = dx$, and $x = t-\alpha$.

Crucially, we must change our limits: when $x=0$, $t=\alpha$, and when $x=\alpha$, $t=2\alpha$. The integral becomes:

$$\int_{\alpha}^{2\alpha} (t-\alpha)\sqrt{t} dt = \int_{\alpha}^{2\alpha} (t^{3/2} - \alpha t^{1/2}) dt$$

Integrating this term by term gives us:

$$\left[ \frac{2}{5}t^{5/2} - \frac{2}{3}\alpha t^{3/2} \right]_{\alpha}^{2\alpha}$$

### The Final Symmetry

After carefully evaluating the limits and combining the results, we find that the integral simplifies to:

$$I = \frac{\alpha^{3/2}}{15} (4\sqrt{2} + 10)$$

Now, we equate this to the given value $\frac{16+20\sqrt{2}}{15}$. The denominators cancel out, leaving us with:

$$\alpha^{3/2} (4\sqrt{2} + 10) = 16+20\sqrt{2}$$

Look closely at the right side. We can factor out $4\sqrt{2}$ to get $4\sqrt{2}(2\sqrt{2} + 5)$. On the left, we factor out $2$ to get $2\alpha^{3/2}(2\sqrt{2} + 5)$.

The term $(2\sqrt{2}+5)$ cancels out perfectly on both sides. We are left with:

$$\alpha^{3/2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$

Since $2\sqrt{2} = 2^{3/2}$, we conclude that the final answer is:

$$\mathbf{\alpha = 2}$$

Animated Solution for Mathematics - Definite Integration: Let $α > 0$ . If $\int_{0}^{α} \frac{x}{x + α - x} d x = \frac{16 + 20 2}{15}$ , then $α$ is equal to :

Select Answer:

Visualized Solution

Problem Introduction

Rationalizing the Denominator

Simplifying the Integral

Integrating $x^{3/2}$

Substitution for the First Part

Integrating the Substituted Expression

Evaluating the Limits

Combining the Results

Equating to the Given Value

Solving for $α$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Art of Rationalization

Divide and Conquer

The Substitution

The Final Symmetry

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