JEE PYQ: Indefinite Integration Question 11
Question 11 - 2020 (05 Sep Shift 1)
If $\int (e^{2x} + 2e^x - e^{-x} - 1)e^{(e^x + e^{-x})},dx = g(x)e^{(e^x + e^{-x})} + c$, where $c$ is a constant of integration, then $g(0)$ is equal to:
(1) $e$
(2) $e^2$
(3) 1
(4) 2
Show Answer
Answer: (4)
Solution
Let $e^x + e^{-x} + x = t$, then $(e^x - e^{-x} + 1),dx = dt$. Rewrite: $I = \int (e^x - e^{-x} + 1)e^{(e^x+e^{-x})}\cdot e^x,dx + \int (e^x - e^{-x})e^{(e^x+e^{-x})},dx$. Using $\int e^x(f(x) + f’(x)),dx = e^x f(x)$ pattern: $g(x) = 1 + e^x$. So $g(0) = 1 + 1 = 2$.
BETA
