Prove
Question:

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

(A) $10^{x}-x^{10}+\mathrm{C}$

(B) $10^{x}+x^{10}+\mathrm{C}$

(C) $\left(10^{x}-x^{10}\right)^{-1}+\mathrm{C}$

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

Solution:

Let $x^{10}+10^{x}=t$

$\therefore\left(10 x^{9}+10^{x} \log _{e} 10\right) d x=d t$

$\Rightarrow \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \frac{d t}{t}$

$=\log t+\mathrm{C}$

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

Hence, the correct answer is D.

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