allery$\int \dfrac{10x^9+10^{x}\log_e 10}{x^{10}+10^{x}}\,dx$ equals
A$10^{x}-x^{10}+C$
B$10^{x}+x^{10}+C$
C$\left(10^{x}-x^{10}\right)^{-1}+C$
D$\log\!\left(10^{x}+x^{10}\right)+C$
Answer & Solution
Correct answer: D. $\log\!\left(10^{x}+x^{10}\right)+C$
1. Let $t=x^{10}+10^{x}$.
2. Then $\dfrac{dt}{dx}=10x^9+10^{x}\log_e 10$, exactly the numerator.
3. So the integral is $\int \dfrac{dt}{t}=\log|t|$.
4. Hence $\log\!\left(x^{10}+10^{x}\right)+C$. Options A and B mistake the numerator for an antiderivative.
_Source: syllabus Mathematics Ch 7 "Integrals", p.304_
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