Definite Integration Question 409
Question: If $ x(x^{4}+1)\varphi (x)=1, $ then $ \int_1^{2}{\varphi (x)dx=} $
[SCRA 1986]
Options:
A) $ \frac{1}{4}\log \frac{32}{17} $
B) $ \frac{1}{2}\log \frac{32}{17} $
C) $ \frac{1}{4}\log \frac{16}{17} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Here $ \varphi (x)=\frac{1}{x(x^{4}+1)}=\frac{1}{x}-\frac{x^{3}}{x^{4}+1} $
therefore $ \int_1^{2}{\varphi (x)dx=\int_1^{2}{( \frac{1}{x}-\frac{x^{3}}{x^{4}+1} )}dx} $
$ =|\log x|_1^{2}-| \frac{1}{4}\log (x^{2}+1) |_1^{2}=\frac{1}{4}\log \frac{32}{17} $ .
BETA
