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Q.

The value of ∫dx5+4cos⁡x 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

answer is D.

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Detailed Solution

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