int frac{10 x^{9}+10^{x} log _{e} 10}{x^{10}+ | vedclass

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(D) Let $t = x^{10} + 10^{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^{x})$$\fra

Vedclass · Accepted Answer

$\log \left(10^{x}+x^{10}\right)+C$

Vedclass · Answer

(D) Let $t = x^{10} + 10^{x}$.Differentiating both sides with respect to $x$,we get:$\frac{dt}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(10^{x})$$\frac{dt}{dx} = 10x^{9} + 10^{x} \log_{e} 10$Therefore,$dt = (10x^{9} + 10^{x} \log_{e} 10) dx$.Substituting these into the integral,we have:$\int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x = \int \frac{1}{t} dt$$= \log |t| + C$$= \log |x^{10} + 10^{x}| + C$Since $x^{10} + 10^{x} > 0$ for all real $x$,we can write this as $\log (x^{10} + 10^{x}) + C$.Hence,the correct option is $D$.

$\int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x$ equals

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