integral of tanx/logsecx
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Answer:
ANSWER
∫e
x
(tanx+logsecx)dx
=∫e
x
tanxdx+∫e
x
logsecxdx
=∫e
x
tanxdx+log(secx)e
x
−∫
dx
d
(logsecx).∫e
x
dx
=∫e
x
tanxdx+log(sec)xe
x
−∫e
x
tanxdx
=e
x
logsecx+c
Step-by-step explanation:
hope it helpfull for you
THANKYOU
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