Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 391 Question 391 of 416 Hard MCQ
If
\[\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} \] = \[g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c\]
where c is a constant of integration,
then g(0) is equal to :
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I = \[\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} \]
= \[\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx\]
= \[\int {\left( {{e^{2x}} + {e^x} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx\]
+ \[\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} \]
= \[\int {{{\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \over {{e^x}}}.} {e^x}dx\]
\[ + \int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} \]
= \[\int {\left( {{e^x} - {e^{ - x}} + 1} \right)} {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx\]
+ \[\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} \]
Let e x + e − x + x {{e^x} + {e^{ - x}} + x} e x + e − x + x = t \[ \Rightarrow \] \[{\left( {{e^x} - {e^{ - x}} + 1} \right)}\]dx = dt
and let e x + e − x {{e^x} + {e^{ - x}}} e x + e − x = u \[ \Rightarrow \] \[{\left( {{e^x} - {e^{ - x}}} \right)dx}\] = du
\[ \therefore \] I = \[\int {{e^t}} dt + \int {{e^u}} du\]
= e t {{e^t}} e t + e u {{e^u}} e u + C
= \[{e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}\] + \[{{e^{\left( {{e^x} + {e^{ - x}}} \right)}}}\] + C
= \[{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right)\] + C
\[ \therefore \] g(x) = ex + 1
\[ \Rightarrow \] g(0) = 2 Answer the question to unlock