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a
122tan−13tan x+122+C
b
122tan−1tan x+122+C
c
162tan−13tan x+122+C
d
12tan−13tan x+122+C
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Correct option is A
I=∫13+sin2xdx=∫13sin2x+cos2x+2sinxcosxdx=∫sec2x3tan2x+2tanx+3dx[Dividing Nr and Dr by cos2 x] Putting tan x = t and sec2 x dx = dt, we get I=∫dt3t2+2t+3=13∫dtt2+23t+1=13∫dtt+132+2232I=131223tan−1t+13223+C=122tan−13t+122+C=122tan−13tan x+122+CSimilar Questions
is equal to
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