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Maths Question 18 – JEE-MAIN 2025

4∫011(3+x2+1+x2)dx−3loge⁡(3) is equal to :

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

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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.

Step 1: Rationalize the Integrand✦ Active

Rationalize the denominator of the integrand by multiplying the numerator and denominator by the conjugate of the denominator, (3+x2−1+x2). This simplifies the integrand to a difference of square roots.

13+x2+1+x2=3+x2−1+x2(3+x2)2−(1+x2)2=3+x2−1+x2(3+x2)−(1+x2)=3+x2−1+x22
Step 2: Evaluate the Definite Integral○ Expand

Substitute the simplified integrand back into the integral and evaluate each term using the standard formula ∫a2+x2dx=x2a2+x2+a22loge⁡|x+a2+x2|. Apply the limits from 0 to 1.

I=12[∫013+x2dx−∫011+x2dx] ∫013+x2dx=[x23+x2+32loge⁡|x+3+x2|]01=(1+32loge⁡(3))−32loge⁡(3)=1+34loge⁡(3) ∫011+x2dx=[x21+x2+12loge⁡|x+1+x2|]01=22+12loge⁡(1+2) I=12[(1+34loge⁡(3))−(22+12loge⁡(1+2))]
💡 Teacher's Secret Hint

Remember to simplify logarithmic terms using properties like loge⁡(ab)=bloge⁡(a).

Step 3: Calculate the Final Expression○ Expand

Substitute the value of I into the given expression 4I−3loge⁡(3) and simplify. Note that 3loge⁡(3)=32loge⁡(3).

4I=4⋅12[1+34loge⁡(3)−22−12loge⁡(1+2)] 4I=2+32loge⁡(3)−2−loge⁡(1+2) 4I−3loge⁡(3)=(2+32loge⁡(3)−2−loge⁡(1+2))−32loge⁡(3) =2−2−loge⁡(1+2)
💡 Teacher's Secret Hint

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

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