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$\int (\frac{10 x^{9} + 10^{x} \log_{e} 10}{x^{10} + 10^{x}}) dx$ is equal to

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10x+x10+C

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log⁡10x+x10+C

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detailed solution

Correct option is D

∫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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∫10x9+10xloge⁡10x10+10xdx is equal to

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$\int (\frac{10 x^{9} + 10^{x} \log_{e} 10}{x^{10} + 10^{x}}) dx$ is equal to