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2y4y+36dy is equal to 

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a
log2y+4y+36+k
b
−1log2log2y+4y+36+k
c
1log22y+4y+36+k
d
1log2log2y+2y+36+k

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detailed solution

Correct option is C

∫2y4y+36dy=∫2ydy2y2+36 Put 2y=t2ylog⁡2dy=dt=1log⁡2∫dtt2+62=1log⁡2log⁡t+t2+62+k=1log⁡2log⁡2y+4y+36+k


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