4 integral_0^1 (1 / (sqrt(3+x^2) + sqrt(1+x^2))) dx - 3log_e(sqrt(3)) is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2025

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Home Maths IntegralsQuestion #215

4 \int_{0}^{1} \frac{1}{(\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}})} d x - 3 \log_{e} (\sqrt{3})

is equal to :

Clue 1 of 3: Simplify the Integrand

Begin by rationalizing the denominator of the integrand to simplify the expression before integration.

🥷

Ninja Strategy•Estimate the Sign of the Result

By roughly estimating the value of the integral and the logarithmic term, one can deduce that the final expression is likely negative, eliminating all positive options.

🧠 Full Solution Path

Step 1: Rationalize the Integrand✦ Active

Rationalize the denominator of the integrand by multiplying the numerator and denominator by the conjugate of the denominator, $(\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}})$ . This simplifies the integrand to a difference of square roots.

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

Step 2: Evaluate the Definite Integral○ Expand

Substitute the simplified integrand back into the integral and evaluate each term using the standard formula $\int \sqrt{a^{2} + x^{2}} d x = \frac{x}{2} \sqrt{a^{2} + x^{2}} + \frac{a^{2}}{2} \log_{e} | x + \sqrt{a^{2} + x^{2}} |$ . Apply the limits from 0 to 1.

I = \frac{1}{2} [\int_{0}^{1} \sqrt{3 + x^{2}} d x - \int_{0}^{1} \sqrt{1 + x^{2}} d x]

\int_{0}^{1} \sqrt{3 + x^{2}} d x = {[\frac{x}{2} \sqrt{3 + x^{2}} + \frac{3}{2} \log_{e} | x + \sqrt{3 + x^{2}} |]}_{0}^{1} = (1 + \frac{3}{2} \log_{e} (3)) - \frac{3}{2} \log_{e} (\sqrt{3}) = 1 + \frac{3}{4} \log_{e} (3)

\int_{0}^{1} \sqrt{1 + x^{2}} d x = {[\frac{x}{2} \sqrt{1 + x^{2}} + \frac{1}{2} \log_{e} | x + \sqrt{1 + x^{2}} |]}_{0}^{1} = \frac{\sqrt{2}}{2} + \frac{1}{2} \log_{e} (1 + \sqrt{2})

I = \frac{1}{2} [(1 + \frac{3}{4} \log_{e} (3)) - (\frac{\sqrt{2}}{2} + \frac{1}{2} \log_{e} (1 + \sqrt{2}))]

💡 Teacher's Secret Hint

Remember to simplify logarithmic terms using properties like $\log_{e} (a^{b}) = b \log_{e} (a)$ .

Step 3: Calculate the Final Expression○ Expand

Substitute the value of $I$ into the given expression $4 I - 3 \log_{e} (\sqrt{3})$ and simplify. Note that $3 \log_{e} (\sqrt{3}) = \frac{3}{2} \log_{e} (3)$ .

4 I = 4 \cdot \frac{1}{2} [1 + \frac{3}{4} \log_{e} (3) - \frac{\sqrt{2}}{2} - \frac{1}{2} \log_{e} (1 + \sqrt{2})]

4 I = 2 + \frac{3}{2} \log_{e} (3) - \sqrt{2} - \log_{e} (1 + \sqrt{2})

4 I - 3 \log_{e} (\sqrt{3}) = (2 + \frac{3}{2} \log_{e} (3) - \sqrt{2} - \log_{e} (1 + \sqrt{2})) - \frac{3}{2} \log_{e} (3)

= 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2})

💡 Teacher's Secret Hint

Ensure all terms are correctly combined and cancelled, especially the logarithmic terms.

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Maths Question 18 – JEE-MAIN 2025

Clue 1 of 3: Simplify the Integrand

🧠 Full Solution Path