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Q.

∫10x9+10xloge⁡10x10+10xdx is equal to

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a

10x+x10+C

b

10x-x10+C

c

10x−x10−1+C

d

log⁡10x+x10+C

answer is D.

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Detailed Solution

∫10x9+10xloge⁡10x10+10xdxLet x10+10x=t⇒10x9+10xloge⁡10dx=dt⇒ dx=dt10x9+10xloge⁡10∴ ∫10x9+10xloge⁡10x10+10xdx=∫10x9+10xloge⁡10t×dt10x2+10xloge⁡10=∫dtt=log⁡|t|+C=log⁡10x+x10+C
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