4 ∫₀¹ (1/(√(3+x²)+√(1+x²))) dx - 3 logₑ (√3) is equal to :

Q: Wait, why do we rationalize? Can't we just use substitution?

You could try trigonometric substitution (like $x = \tan \theta$), but that would lead to a much more complex integral involving secant powers. Rationalization is the most direct path here because it immediately transforms the denominator into a constant, turning a complex fraction into a simple difference of two standard integrals.

Q: Why does the $3 \ln \sqrt{3}$ term appear and then disappear?

That is the beauty of this problem! The $3 \ln \sqrt{3}$ term is part of the evaluation of the first integral at the lower limit. When you combine all the terms from the integration and subtract the final $3 \ln \sqrt{3}$ as required by the original expression, they cancel out perfectly. It is a designed feature of the problem to test your patience and precision.

Q: Is it possible to make a sign error when splitting the integral?

Absolutely, and that is the most common trap! When you split the integral, ensure the negative sign from the rationalization is distributed correctly to both parts. Always keep the constant factor (in this case, $2$) outside the entire bracket until you have fully evaluated both integrals.

Question

$4 \int_0^1 (\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}) dx - 3 \log_e (\sqrt{3})$ is equal to :

Options:

(A) $2 + \sqrt{2} + \log_e(1+\sqrt{2})$

(B) $2 - \sqrt{2} - \log_e(1+\sqrt{2})$

(C) $2 + \sqrt{2} - \log_e(1+\sqrt{2})$

(D) $2 - \sqrt{2} + \log_e(1+\sqrt{2})$

Sigma Percentile · Accepted Answer

### The Art of Rationalization: Taming the Monster Integral

Have you ever looked at an integral and felt a shiver down your spine? When you first encounter an expression like

$$\int_{0}^{1} \frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}} dx$$

it is natural to feel intimidated. The denominator is a mess of square roots, and there is no obvious substitution that makes it vanish.

But in the world of JEE Advanced, these problems are not designed to defeat you; they are designed to reward the student who knows how to look beneath the surface.

### Phase 1: The Rationalization Strategy

When you see a sum of square roots in the denominator, your mathematical instinct should scream one word: **rationalization**. We are not going to fight the roots; we are going to clear them.

We multiply the numerator and the denominator by the conjugate of the denominator: $\sqrt{3+x^2} - \sqrt{1+x^2}$.

Why? Because of the beautiful algebraic identity $(a+b)(a-b) = a^2 - b^2$. When we apply this to our denominator, the square roots vanish:

$$(\sqrt{3+x^2})^2 - (\sqrt{1+x^2})^2 = (3+x^2) - (1+x^2) = 2$$

Suddenly, the denominator is just a constant, $2$. Our terrifying integral has collapsed into a much friendlier form:

$$\int_{0}^{1} \frac{\sqrt{3+x^2}-\sqrt{1+x^2}}{2} dx = \frac{1}{2} \int_{0}^{1} (\sqrt{3+x^2} - \sqrt{1+x^2}) dx$$

### Phase 2: The Geometric Perspective

Geometrically, what are we doing here? We are calculating the area trapped between two curves, $y = \sqrt{3+x^2}$ and $y = \sqrt{1+x^2}$, from $x=0$ to $x=1$.

By splitting the integral into two parts, we are essentially finding the area under the first curve and subtracting the area under the second. This visual intuition is what separates a calculator from a physicist.

### Phase 3: The Heavy Lifting

Now, we need our most reliable tool: the standard integral formula for $\int \sqrt{a^2+x^2} dx$. This formula is a cornerstone of JEE calculus:

$$\int \sqrt{a^2+x^2} dx = \frac{x}{2}\sqrt{a^2+x^2} + \frac{a^2}{2} \ln|x+\sqrt{a^2+x^2}|$$

For our first integral, $a^2 = 3$. For the second, $a^2 = 1$. We apply the limits $0$ to $1$ to both.

For the first integral, the upper limit $x=1$ gives us $\frac{1}{2}\sqrt{4} + \frac{3}{2} \ln(1+2) = 1 + \frac{3}{2} \ln 3$, and the lower limit $x=0$ gives us $\frac{3}{2} \ln \sqrt{3}$.

For the second integral, the upper limit gives us $\frac{1}{2}\sqrt{2} + \frac{1}{2} \ln(1+\sqrt{2})$, and the lower limit is $\frac{1}{2} \ln(1) = 0$.

### Phase 4: The Logarithmic Dance

Now, we combine everything. We have:

$$\frac{1}{2} \left[ \left( 1 + \frac{3}{2} \ln 3 - \frac{3}{2} \ln \sqrt{3} \right) - \left( \frac{\sqrt{2}}{2} + \frac{1}{2} \ln(1+\sqrt{2}) \right) \right]$$

Distributing the $\frac{1}{2}$, we simplify the expression. Note that $3 \ln 3 = 6 \ln \sqrt{3}$, so $3 \ln 3 - 3 \ln \sqrt{3} = 3 \ln \sqrt{3}$.

After careful algebraic reduction, the logarithmic terms consolidate into a final form. We are left with the elegant result:

$$\frac{2 - \sqrt{2} + 3 \ln \sqrt{3} - \ln(1+\sqrt{2})}{2}$$

### Conclusion: The JEE Mindset

This problem was never about brute force. It was about recognizing a pattern, applying the right tool, and having the confidence to trust the process until the final cancellation.

Keep this mindset, and you will find that even the most intimidating integrals are just puzzles waiting to be solved.

Animated Solution for Mathematics - Definite Integration: $4 \int_{0}^{1} (\frac{1}{3 + x ^{2} + 1 + x ^{2}}) d x - 3 lo g_{e} (3)$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Integrand

The Strategy: Rationalization

Simplifying the Denominator

Splitting the Integral

Visualizing the Curves

The Standard Formula

Applying Limits to the First Term

Applying Limits to the Second Term

Combining the Results

Simplifying Logarithmic Terms

Final Subtraction and Answer

The Way Forward

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Art of Rationalization

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4∫01​(3+x2​+1+x2​1​)dx−3loge​(3​) is equal to :