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∫e^x(tanx+1)secx
=∫e^x(secx.tanx + secx)
it is of form
∫e^x(f(x) + f'(x))
=e^x.f(x) + C
here, f(x) = secx
hence,
I = e^x.f(x) + C
= e^xsecx + C
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Answer:
∫e^x(tanx+1)secx
=∫e^x(secx.tanx + secx)
it is of form
∫e^x(f(x) + f'(x))
=e^x.f(x) + C
here, f(x) = secx
hence,
I = e^x.f(x) + C
= e^xsecx + C
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