PYQs Homework Notes Mocks Board Daily Homework MHT-CET Maths Question type Integrals with x e^x: substitute t = x e^x This type of question, asked 5 times: 9 May 2023 S2, 11 May 2023 S2, 9 May 2024 S1, 13 Apr 2026 S2, 18 Apr 2026 S1. All 5 questions below, oldest first.
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Question 1 9 May 2023 S2
102 The value of ∫ x + 1 x ( 1 + x e x ) 2 d x \int \frac{x+1}{x(1+xe^x)^2}\,dx ∫ x ( 1 + x e x ) 2 x + 1 d x is equal to A. log x e x 1 + x e x + x 1 + x e x + c \log\frac{xe^x}{1+xe^x} + \frac{x}{1+xe^x} + c log 1 + x e x x e x + 1 + x e x x + c B. log x e x 1 + x e x + e x 1 + x e x + c \log\frac{xe^x}{1+xe^x} + \frac{e^x}{1+xe^x} + c log 1 + x e x x e x + 1 + x e x e x + c C. log x e x 1 + x e x + 1 1 + x e x + c \log\frac{xe^x}{1+xe^x} + \frac{1}{1+xe^x} + c log 1 + x e x x e x + 1 + x e x 1 + c D. log x e x 1 + x e x − x 1 + x e x + c \log\frac{xe^x}{1+xe^x} - \frac{x}{1+xe^x} + c log 1 + x e x x e x − 1 + x e x x + c Tap an option to check your answer.
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Question 2 11 May 2023 S2
103 ∫ x + 1 x ( 1 + x e x ) 2 d x = \int\frac{x+1}{x(1+xe^x)^2}\,dx = ∫ x ( 1 + x e x ) 2 x + 1 d x = A. log ∣ x e x 1 + x e x ∣ + x 1 + x e x + c \log\left|\frac{xe^x}{1+xe^x}\right|+\frac{x}{1+xe^x}+c log 1 + x e x x e x + 1 + x e x x + c B. log ∣ x e x 1 + x e x ∣ + e x 1 + x e x + c \log\left|\frac{xe^x}{1+xe^x}\right|+\frac{e^x}{1+xe^x}+c log 1 + x e x x e x + 1 + x e x e x + c C. log ∣ x e x 1 + x e x ∣ + 1 1 + x e x + c \log\left|\frac{xe^x}{1+xe^x}\right|+\frac{1}{1+xe^x}+c log 1 + x e x x e x + 1 + x e x 1 + c D. log ∣ x e x 1 + x e x ∣ − 1 1 + x e x + c \log\left|\frac{xe^x}{1+xe^x}\right|-\frac{1}{1+xe^x}+c log 1 + x e x x e x − 1 + x e x 1 + c Tap an option to check your answer.
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Question 3 9 May 2024 S1
135 ∫ ( x + 1 ) x ( 1 + x e x ) 2 d x = \int\frac{(x+1)}{x(1+xe^x)^2}\,dx = ∫ x ( 1 + x e x ) 2 ( x + 1 ) d x = A A. log x e x 1 + x e x + c \log\frac{xe^x}{1+xe^x}+c log 1 + x e x x e x + c B. log x e x 1 + x e x − 1 1 + x e x + c \log\frac{xe^x}{1+xe^x}-\frac{1}{1+xe^x}+c log 1 + x e x x e x − 1 + x e x 1 + c C. log 1 + x e x + 1 1 + x e x + c \log\frac{1+xe^x}+\frac{1}{1+xe^x}+c log + 1 + x e x 1 + x e x 1 + c D. log x e x 1 + x e x + 1 1 + x e x + c \log\frac{xe^x}{1+xe^x}+\frac{1}{1+xe^x}+c log 1 + x e x x e x + 1 + x e x 1 + c Tap an option to check your answer.
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Question 4 13 Apr 2026 S2
78 ∫ ( x + 1 ) x ( 1 + x e x ) 2 d x = \displaystyle\int\dfrac{(x + 1)}{x(1 + xe^x)^2}\,dx = ∫ x ( 1 + x e x ) 2 ( x + 1 ) d x = A. − log ( x e x 1 + x e x ) + 1 ( 1 + x e x ) + c -\log\left(\dfrac{xe^x}{1 + xe^x}\right) + \dfrac{1}{(1 + xe^x)} + c − log ( 1 + x e x x e x ) + ( 1 + x e x ) 1 + c B. log ( x e x 1 + x e x ) + 1 ( 1 + x e x ) + c \log\left(\dfrac{xe^x}{1 + xe^x}\right) + \dfrac{1}{(1 + xe^x)} + c log ( 1 + x e x x e x ) + ( 1 + x e x ) 1 + c C. log ( 1 + x e x x e x ) + ( 1 + x e x ) + c \log\left(\dfrac{1 + xe^x}{xe^x}\right) + (1 + xe^x) + c log ( x e x 1 + x e x ) + ( 1 + x e x ) + c D. − log ( x e x 1 + x e x ) − 1 ( 1 + x e x ) + c -\log\left(\dfrac{xe^x}{1 + xe^x}\right) - \dfrac{1}{(1 + xe^x)} + c − log ( 1 + x e x x e x ) − ( 1 + x e x ) 1 + c Tap an option to check your answer.
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Question 5 18 Apr 2026 S1
81 ∫ ( x + 1 ) d x x ( 1 + x e x ) = … … \displaystyle\int\dfrac{(x + 1)\,dx}{x(1 + xe^x)} = \ldots\ldots ∫ x ( 1 + x e x ) ( x + 1 ) d x = …… A. log ∣ x e x 1 + x e x ∣ + c \log\left|\dfrac{xe^x}{1 + xe^x}\right| + c log 1 + x e x x e x + c B. log ∣ 1 + x e x x e x ∣ + c \log\left|\dfrac{1 + xe^x}{xe^x}\right| + c log x e x 1 + x e x + c C. log ∣ ( x + 1 ) e x x e x ∣ + c \log\left|\dfrac{(x + 1)e^x}{xe^x}\right| + c log x e x ( x + 1 ) e x + c D. log ∣ ( x + 1 ) e x 1 + x e x ∣ + c \log\left|\dfrac{(x + 1)e^x}{1 + xe^x}\right| + c log 1 + x e x ( x + 1 ) e x + c Tap an option to check your answer.
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