Integral Calculus Question 71
Question: $ \int_{{}}^{{}}{\frac{x^{2}+1}{x^{4}+1}dx=} $
[AISSE 1990]
Options:
A) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{x^{2}-1}{2x} )+c $
B) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{x^{2}-1}{\sqrt{2x}} )+c $
C) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{x^{2}-1}{2\sqrt{x}} )+c $
D) $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{x^{2}-1}{\sqrt{2}x} )+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{x^{2}+1}{x^{4}+1}},dx=\int_{{}}^{{}}{\frac{( 1+\frac{1}{x^{2}} )}{( x^{2}+\frac{1}{x^{2}} )}},dx=\int_{{}}^{{}}{\frac{( 1+\frac{1}{x^{2}} ),dx}{{{( x-\frac{1}{x} )}^{2}}+2}} $ Put $ x-\frac{1}{x}=t\Rightarrow ( 1+\frac{1}{x^{2}} ),dx=dt, $ then the required integral is $ \frac{1}{\sqrt{2}}{{\tan }^{-1}}( \frac{x^{2}-1}{\sqrt{2}x} )+c $ .
BETA
