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MathematicsIndefinite IntegrationBy partHard2 minPYQ_2020
MathematicsHardsingle choice

Ife2x+2ex-e-x-1eex+e-xdx=g(x)eex+e-x+c,where c is a constant of integration, theng(0)is

Options:

Answer:
D
Solution:

I=e2x+2ex-e-x-1eex+e-xdx

I=e2x+ex-1eex+e-xdx+ex-e-xeex+e-xdx

I=ex+1-e-xeex+e-x+xdx+eex+e-x+c

ex+e-x+x=u

ex-e-x+1dx=du

I=eex+e-x+x+eex+e-x+c=eex+e-xex+1+c

then g(x)=ex+1

g(0)=2

Stream:JEESubject:MathematicsTopic:Indefinite IntegrationSubtopic:By part
2mℹ️ Source: PYQ_2020

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