Integral Calculus Question 402
Question: $ \int_{{}}^{{}}{\frac{10x^{9}+10^{x}{\log_{e}}10}{10^{x}+x^{10}}}\ dx= $
[MNR 1979]
Options:
A) $ -\frac{1}{2}\frac{1}{{{(10^{x}+x^{10})}^{2}}}+c $
B) $ \log (10^{x}+x^{10})+c $
C) $ \frac{1}{2}\frac{1}{{{(10^{x}+x^{10})}^{2}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ x^{10}+10^{x}=t\Rightarrow (10x^{9}+10^{x}{\log_{e}}10),dx=dt, $ then $ \int_{{}}^{{}}{\frac{10x^{9}+10^{x}{\log_{e}}10}{10^{x}+x^{10}},dx}=\int_{{}}^{{}}{\frac{1}{t},dt=\log t+c} $ $ =\log (x^{10}+10^{x})+c. $
BETA
