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The value of dx5+4cosx is

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a
13tan−1⁡13tan⁡x+C
b
13tan−1⁡13tan⁡x2+C
c
23tan−1⁡13tan⁡x+C
d
23tan−1⁡13tan⁡x2+C

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detailed solution

Correct option is D

∫dx5+4cos⁡x=∫2dt51+t2+41−t2(t=tan⁡(x/2))=2∫dt9+t2=23tan−1⁡13tan⁡x2+C


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