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.