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Integral Calculus: JEE Main Mathematics Question with Solution The value of the integral
∫ − log e 2 log e 2 e x ( log e ( e x + 1 + e 2 x ) ) d x \int_{-\log_e 2}^{\log_e 2} e^x \left( \log_e \left( e^x + \sqrt{1 + e^{2x}} \right) \right) dx ∫ − l o g e 2 l o g e 2 e x ( log e ( e x + 1 + e 2 x ) ) d x is equal to
Hint 1 of 3
What is the most direct substitution to simplify the integral I = ∫ − log 2 log 2 e x log ( e x + 1 + e 2 x ) d x I = \int_{-\log 2}^{\log 2} e^x \log\left(e^x + \sqrt{1 + e^{2x}}\right) dx I = ∫ − l o g 2 l o g 2 e x log ( e x + 1 + e 2 x ) d x
Substitute t = e x t = e^x t = e x d t = e x d x dt = e^x dx d t = e x d x 1 / 2 1/2 1/2 2 2 2 Substitute u = e x + 1 + e 2 x u = e^x + \sqrt{1+e^{2x}} u = e x + 1 + e 2 x 1 + 5 2 \frac{1+\sqrt{5}}{2} 2 1 + 5 2 + 5 2+\sqrt{5} 2 + 5 Step-by-step solution View Correct answer
Using the substitution t = e x t = e^x t = e x log e ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 \log_e \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5 Option analysis
Why each option works or fails A · log e ( 2 ( 2 + 5 ) 1 + 5 ) − 5 2 \log_e \left( \frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) ) − 2 5 You miscalculated the coefficient when factoring out power terms inside the logarithm, replacing 2 \sqrt{2} 2 2 2 2 ( 2 + 5 ) (2+\sqrt{5}) ( 2 + 5 ) 2 log ( 2 + 5 ) − 1 2 log ( 1 / 2 + 5 / 2 ) 2\log(2+\sqrt{5}) - \frac{1}{2}\log(1/2+\sqrt{5}/2) 2 log ( 2 + 5 ) − 2 1 log ( 1/2 + 5 /2 ) log ( ( 2 + 5 ) 2 / ( 1 + 5 ) / 2 ) = log ( 2 ( 2 + 5 ) 2 1 + 5 ) \log((2+\sqrt{5})^2 / \sqrt{(1+\sqrt{5})/2}) = \log\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right) log (( 2 + 5 ) 2 / ( 1 + 5 ) /2 ) = log ( 1 + 5 2 ( 2 + 5 ) 2 )
B · log e ( 2 ( 3 − 5 ) 2 1 + 5 ) + 5 2 \log_e \left( \frac{\sqrt{2}(3-\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) + \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 3 − 5 ) 2 ) + 2 5 You flipped the sign of the algebraic term during integration by parts from − ∫ v d u -\int v\,du − ∫ v d u ( 2 + 5 ) (2+\sqrt{5}) ( 2 + 5 ) ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u [ 1 + t 2 ] 1 / 2 2 [\sqrt{1+t^2}]_{1/2}^2 [ 1 + t 2 ] 1/2 2 5 − 5 2 = 5 2 \sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2} 5 − 2 5 = 2 5
C · log e ( ( 2 + 5 ) 2 1 + 5 ) + 5 2 \log_e \left( \frac{(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) + \frac{\sqrt{5}}{2} log e ( 1 + 5 ( 2 + 5 ) 2 ) + 2 5 You forgot the factor of 2 − 1 / 2 = 1 / 2 2^{-1/2} = 1/\sqrt{2} 2 − 1/2 = 1/ 2 t = 1 / 2 t = 1/2 t = 1/2 1 + t 2 = 5 / 2 \sqrt{1+t^2} = \sqrt{5}/2 1 + t 2 = 5 /2 t + 1 + t 2 = 1 + 5 2 t + \sqrt{1+t^2} = \frac{1+\sqrt{5}}{2} t + 1 + t 2 = 2 1 + 5 1 / 2 1/2 1/2 2 \sqrt{2} 2
D · log e ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 \log_e \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5 None. This is the correct value of the definite integral. Perform substitution t = e x t = e^x t = e x u = log ( t + 1 + t 2 ) u = \log(t+\sqrt{1+t^2}) u = log ( t + 1 + t 2 ) d v = d t dv = dt d v = d t t ∈ [ 1 / 2 , 2 ] t \in [1/2, 2] t ∈ [ 1/2 , 2 ]
Step Working
01 given Integral I = ∫ − ln 2 ln 2 e x ln ( e x + 1 + e 2 x ) d x I = \int_{-\ln 2}^{\ln 2} e^x \ln\left(e^x + \sqrt{1 + e^{2x}}\right) dx I = ∫ − l n 2 l n 2 e x ln ( e x + 1 + e 2 x ) d x
02 approach Substitute t = e x t = e^x t = e x d t = e x d x dt = e^x dx d t = e x d x x = − ln 2 x = -\ln 2 x = − ln 2 t = 1 / 2 t = 1/2 t = 1/2 x = ln 2 x = \ln 2 x = ln 2 t = 2 t = 2 t = 2 ∫ 1 / 2 2 ln ( t + 1 + t 2 ) d t \int_{1/2}^2 \ln(t + \sqrt{1+t^2}) dt ∫ 1/2 2 ln ( t + 1 + t 2 ) d t
03 execute Let t = e x ⟹ d t = e x d x t = e^x \implies dt = e^x dx t = e x ⟹ d t = e x d x t = e − ln 2 = 1 / 2 t = e^{-\ln 2} = 1/2 t = e − l n 2 = 1/2 t = e ln 2 = 2 t = e^{\ln 2} = 2 t = e l n 2 = 2 I = ∫ 1 / 2 2 ln ( t + 1 + t 2 ) d t I = \int_{1/2}^2 \ln\left(t + \sqrt{1+t^2}\right) dt I = ∫ 1/2 2 ln ( t + 1 + t 2 ) d t ∫ u ⋅ 1 d t = u ⋅ t − ∫ t ⋅ u ′ d t \int u \cdot 1 \, dt = u \cdot t - \int t \cdot u' \, dt ∫ u ⋅ 1 d t = u ⋅ t − ∫ t ⋅ u ′ d t u = ln ( t + 1 + t 2 ) u = \ln(t + \sqrt{1+t^2}) u = ln ( t + 1 + t 2 ) u ′ = 1 1 + t 2 u' = \frac{1}{\sqrt{1+t^2}} u ′ = 1 + t 2 1 I = [ t ln ( t + 1 + t 2 ) ] 1 / 2 2 − ∫ 1 / 2 2 t 1 + t 2 d t I = \left[ t \ln\left(t + \sqrt{1+t^2}\right) \right]_{1/2}^2 - \int_{1/2}^2 \frac{t}{\sqrt{1+t^2}} dt I = [ t ln ( t + 1 + t 2 ) ] 1/2 2 − ∫ 1/2 2 1 + t 2 t d t = 2 ln ( 2 + 5 ) − 1 2 ln ( 1 2 + 1 + 1 / 4 ) = 2 ln ( 2 + 5 ) − 1 2 ln ( 1 + 5 2 ) = 2\ln(2 + \sqrt{5}) - \frac{1}{2}\ln\left(\frac{1}{2} + \sqrt{1 + 1/4}\right) = 2\ln(2 + \sqrt{5}) - \frac{1}{2}\ln\left(\frac{1+\sqrt{5}}{2}\right) = 2 ln ( 2 + 5 ) − 2 1 ln ( 2 1 + 1 + 1/4 ) = 2 ln ( 2 + 5 ) − 2 1 ln ( 2 1 + 5 ) ∫ 1 / 2 2 t 1 + t 2 d t = [ 1 + t 2 ] 1 / 2 2 = 5 − 5 / 4 = 5 − 5 2 = 5 2 \int_{1/2}^2 \frac{t}{\sqrt{1+t^2}} dt = \left[ \sqrt{1+t^2} \right]_{1/2}^2 = \sqrt{5} - \sqrt{5/4} = \sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2} ∫ 1/2 2 1 + t 2 t d t = [ 1 + t 2 ] 1/2 2 = 5 − 5/4 = 5 − 2 5 = 2 5 I = ln ( ( 2 + 5 ) 2 ) − ln ( 5 + 1 2 ) − 5 2 = ln ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 I = \ln\left((2+\sqrt{5})^2\right) - \ln\left(\sqrt{\frac{\sqrt{5}+1}{2}}\right) - \frac{\sqrt{5}}{2} = \ln\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right) - \frac{\sqrt{5}}{2} I = ln ( ( 2 + 5 ) 2 ) − ln ( 2 5 + 1 ) − 2 5 = ln ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5
✓ verify Check that d d t ln ( t + 1 + t 2 ) = 1 + t / 1 + t 2 t + 1 + t 2 = 1 1 + t 2 \frac{d}{dt}\ln(t+\sqrt{1+t^2}) = \frac{1+t/\sqrt{1+t^2}}{t+\sqrt{1+t^2}} = \frac{1}{\sqrt{1+t^2}} d t d ln ( t + 1 + t 2 ) = t + 1 + t 2 1 + t / 1 + t 2 = 1 + t 2 1 [ − ln 2 , ln 2 ] [-\ln 2, \ln 2] [ − ln 2 , ln 2 ]
Hints that build this answer step by step What is the most direct substitution to simplify the integral I = ∫ − log 2 log 2 e x log ( e x + 1 + e 2 x ) d x I = \int_{-\log 2}^{\log 2} e^x \log\left(e^x + \sqrt{1 + e^{2x}}\right) dx I = ∫ − l o g 2 l o g 2 e x log ( e x + 1 + e 2 x ) d x
Substitute t = e x t = e^x t = e x d t = e x d x dt = e^x dx d t = e x d x 1 / 2 1/2 1/2 2 2 2 Applying integration by parts ∫ u d v = u v − ∫ v d u \int u\, dv = uv - \int v\, du ∫ u d v = uv − ∫ v d u ∫ log ( t + 1 + t 2 ) d t \int \log(t + \sqrt{1+t^2}) dt ∫ log ( t + 1 + t 2 ) d t
t log ( t + 1 + t 2 ) − 1 + t 2 t \log(t + \sqrt{1+t^2}) - \sqrt{1+t^2} t log ( t + 1 + t 2 ) − 1 + t 2 Evaluating [ t log ( t + 1 + t 2 ) − 1 + t 2 ] 1 / 2 2 \left[ t \log(t + \sqrt{1+t^2}) - \sqrt{1+t^2} \right]_{1/2}^2 [ t log ( t + 1 + t 2 ) − 1 + t 2 ] 1/2 2
log ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 \log\left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5 Your next move We think you should solve this next ✓ Source and academic review↓
Question type Single correct
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
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Editorial review 8 September 2026 Quick checks
Students also ask Why is the derivative of ln ( t + 1 + t 2 ) \ln(t + \sqrt{1+t^2}) ln ( t + 1 + t 2 ) 1 1 + t 2 \frac{1}{\sqrt{1+t^2}} 1 + t 2 1 By chain rule, d d t ln ( t + 1 + t 2 ) = 1 + 2 t 2 1 + t 2 t + 1 + t 2 = 1 + t 2 + t 1 + t 2 t + 1 + t 2 = 1 1 + t 2 \frac{d}{dt}\ln(t+\sqrt{1+t^2}) = \frac{1 + \frac{2t}{2\sqrt{1+t^2}}}{t+\sqrt{1+t^2}} = \frac{\frac{\sqrt{1+t^2}+t}{\sqrt{1+t^2}}}{t+\sqrt{1+t^2}} = \frac{1}{\sqrt{1+t^2}} d t d ln ( t + 1 + t 2 ) = t + 1 + t 2 1 + 2 1 + t 2 2 t = t + 1 + t 2 1 + t 2 1 + t 2 + t = 1 + t 2 1
Answer Using the substitution t = e x t = e^x t = e x log e ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 \log_e \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5
Why each option works or fails A: log e ( 2 ( 2 + 5 ) 1 + 5 ) − 5 2 \log_e \left( \frac{2(2+\sqrt{5})}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) ) − 2 5 - You miscalculated the coefficient when factoring out power terms inside the logarithm, replacing 2 \sqrt{2} 2 2 2 2 ( 2 + 5 ) (2+\sqrt{5}) ( 2 + 5 ) 2 log ( 2 + 5 ) − 1 2 log ( 1 / 2 + 5 / 2 ) 2\log(2+\sqrt{5}) - \frac{1}{2}\log(1/2+\sqrt{5}/2) 2 log ( 2 + 5 ) − 2 1 log ( 1/2 + 5 /2 ) log ( ( 2 + 5 ) 2 / ( 1 + 5 ) / 2 ) = log ( 2 ( 2 + 5 ) 2 1 + 5 ) \log((2+\sqrt{5})^2 / \sqrt{(1+\sqrt{5})/2}) = \log\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right) log (( 2 + 5 ) 2 / ( 1 + 5 ) /2 ) = log ( 1 + 5 2 ( 2 + 5 ) 2 ) B: log e ( 2 ( 3 − 5 ) 2 1 + 5 ) + 5 2 \log_e \left( \frac{\sqrt{2}(3-\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) + \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 3 − 5 ) 2 ) + 2 5 - You flipped the sign of the algebraic term during integration by parts from − ∫ v d u -\int v\,du − ∫ v d u ( 2 + 5 ) (2+\sqrt{5}) ( 2 + 5 ) ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u [ 1 + t 2 ] 1 / 2 2 [\sqrt{1+t^2}]_{1/2}^2 [ 1 + t 2 ] 1/2 2 5 − 5 2 = 5 2 \sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2} 5 − 2 5 = 2 5 C: log e ( ( 2 + 5 ) 2 1 + 5 ) + 5 2 \log_e \left( \frac{(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) + \frac{\sqrt{5}}{2} log e ( 1 + 5 ( 2 + 5 ) 2 ) + 2 5 - You forgot the factor of 2 − 1 / 2 = 1 / 2 2^{-1/2} = 1/\sqrt{2} 2 − 1/2 = 1/ 2 t = 1 / 2 t = 1/2 t = 1/2 1 + t 2 = 5 / 2 \sqrt{1+t^2} = \sqrt{5}/2 1 + t 2 = 5 /2 t + 1 + t 2 = 1 + 5 2 t + \sqrt{1+t^2} = \frac{1+\sqrt{5}}{2} t + 1 + t 2 = 2 1 + 5 1 / 2 1/2 1/2 2 \sqrt{2} 2 D · correct: log e ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 \log_e \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} log e ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5 - None. This is the correct value of the definite integral. Perform substitution t = e x t = e^x t = e x u = log ( t + 1 + t 2 ) u = \log(t+\sqrt{1+t^2}) u = log ( t + 1 + t 2 ) d v = d t dv = dt d v = d t t ∈ [ 1 / 2 , 2 ] t \in [1/2, 2] t ∈ [ 1/2 , 2 ] Step-by-step solution given: Integral I = ∫ − ln 2 ln 2 e x ln ( e x + 1 + e 2 x ) d x I = \int_{-\ln 2}^{\ln 2} e^x \ln\left(e^x + \sqrt{1 + e^{2x}}\right) dx I = ∫ − l n 2 l n 2 e x ln ( e x + 1 + e 2 x ) d x approach: Substitute t = e x t = e^x t = e x d t = e x d x dt = e^x dx d t = e x d x x = − ln 2 x = -\ln 2 x = − ln 2 t = 1 / 2 t = 1/2 t = 1/2 x = ln 2 x = \ln 2 x = ln 2 t = 2 t = 2 t = 2 ∫ 1 / 2 2 ln ( t + 1 + t 2 ) d t \int_{1/2}^2 \ln(t + \sqrt{1+t^2}) dt ∫ 1/2 2 ln ( t + 1 + t 2 ) d t execute: Let t = e x ⟹ d t = e x d x t = e^x \implies dt = e^x dx t = e x ⟹ d t = e x d x t = e − ln 2 = 1 / 2 t = e^{-\ln 2} = 1/2 t = e − l n 2 = 1/2 t = e ln 2 = 2 t = e^{\ln 2} = 2 t = e l n 2 = 2 I = ∫ 1 / 2 2 ln ( t + 1 + t 2 ) d t I = \int_{1/2}^2 \ln\left(t + \sqrt{1+t^2}\right) dt I = ∫ 1/2 2 ln ( t + 1 + t 2 ) d t ∫ u ⋅ 1 d t = u ⋅ t − ∫ t ⋅ u ′ d t \int u \cdot 1 \, dt = u \cdot t - \int t \cdot u' \, dt ∫ u ⋅ 1 d t = u ⋅ t − ∫ t ⋅ u ′ d t u = ln ( t + 1 + t 2 ) u = \ln(t + \sqrt{1+t^2}) u = ln ( t + 1 + t 2 ) u ′ = 1 1 + t 2 u' = \frac{1}{\sqrt{1+t^2}} u ′ = 1 + t 2 1 I = [ t ln ( t + 1 + t 2 ) ] 1 / 2 2 − ∫ 1 / 2 2 t 1 + t 2 d t I = \left[ t \ln\left(t + \sqrt{1+t^2}\right) \right]_{1/2}^2 - \int_{1/2}^2 \frac{t}{\sqrt{1+t^2}} dt I = [ t ln ( t + 1 + t 2 ) ] 1/2 2 − ∫ 1/2 2 1 + t 2 t d t = 2 ln ( 2 + 5 ) − 1 2 ln ( 1 2 + 1 + 1 / 4 ) = 2 ln ( 2 + 5 ) − 1 2 ln ( 1 + 5 2 ) = 2\ln(2 + \sqrt{5}) - \frac{1}{2}\ln\left(\frac{1}{2} + \sqrt{1 + 1/4}\right) = 2\ln(2 + \sqrt{5}) - \frac{1}{2}\ln\left(\frac{1+\sqrt{5}}{2}\right) = 2 ln ( 2 + 5 ) − 2 1 ln ( 2 1 + 1 + 1/4 ) = 2 ln ( 2 + 5 ) − 2 1 ln ( 2 1 + 5 ) ∫ 1 / 2 2 t 1 + t 2 d t = [ 1 + t 2 ] 1 / 2 2 = 5 − 5 / 4 = 5 − 5 2 = 5 2 \int_{1/2}^2 \frac{t}{\sqrt{1+t^2}} dt = \left[ \sqrt{1+t^2} \right]_{1/2}^2 = \sqrt{5} - \sqrt{5/4} = \sqrt{5} - \frac{\sqrt{5}}{2} = \frac{\sqrt{5}}{2} ∫ 1/2 2 1 + t 2 t d t = [ 1 + t 2 ] 1/2 2 = 5 − 5/4 = 5 − 2 5 = 2 5 I = ln ( ( 2 + 5 ) 2 ) − ln ( 5 + 1 2 ) − 5 2 = ln ( 2 ( 2 + 5 ) 2 1 + 5 ) − 5 2 I = \ln\left((2+\sqrt{5})^2\right) - \ln\left(\sqrt{\frac{\sqrt{5}+1}{2}}\right) - \frac{\sqrt{5}}{2} = \ln\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right) - \frac{\sqrt{5}}{2} I = ln ( ( 2 + 5 ) 2 ) − ln ( 2 5 + 1 ) − 2 5 = ln ( 1 + 5 2 ( 2 + 5 ) 2 ) − 2 5 verify: Check that d d t ln ( t + 1 + t 2 ) = 1 + t / 1 + t 2 t + 1 + t 2 = 1 1 + t 2 \frac{d}{dt}\ln(t+\sqrt{1+t^2}) = \frac{1+t/\sqrt{1+t^2}}{t+\sqrt{1+t^2}} = \frac{1}{\sqrt{1+t^2}} d t d ln ( t + 1 + t 2 ) = t + 1 + t 2 1 + t / 1 + t 2 = 1 + t 2 1 [ − ln 2 , ln 2 ] [-\ln 2, \ln 2] [ − ln 2 , ln 2 ]